# Are there many different power series representation for a given function?

So I have to find the power series representation for $f(x) = \ln (3-x)$.

I attempted the following:

$$\ln(3-x) = \int {- \frac{1}{3-x} dx}$$ $$= - \int { \frac{1}{1-(x-2)} dx}$$ $$= - \int {\sum_{n=0}^{\infty}{(x-2)^n} dx}$$ $$= \sum_{n=0}^{\infty} {\int(x-2)^ndx}$$ $$= \bigg(-\sum_{n=0}^{\infty} \frac{(x-2)^{n+1}}{n+1}\bigg)+K$$

Then if we let $x=2$, then we obtain that $K=0$. Hence the power series representation for $f(x)$ is $-\sum_{n=0}^{\infty} \frac{(x-2)^{n+1}}{n+1}$, where $|x-2|<1$.

However the answer from my lecturer is given as: $$\ln(3)-\sum_{n=1}^{\infty}{\frac{x^n}{n\cdot3^n}}$$

Am I doing a mistake? Or are there many different power series representation for a given function? Any clarification would be highly appreciated.

• It depends where you want to center your power series. Setting a given center, the power series representation is unique (and it exists for an holomorphic function). – Paolo Leonetti Aug 15 '15 at 12:24
• @PaoloLeonetti thanks for your explanation! that makes perfect sense. however, the question does not really specify the center of the power series representation. does that mean that my answer is actually correct as well? – Aaron Aug 15 '15 at 12:36
• In a word, yes :) Ps. How do you justify the exachange of infinite summation and integral? – Paolo Leonetti Aug 15 '15 at 12:36
• @PaoloLeonetti is that because we are allowed to do term-by-term integration? – Aaron Aug 15 '15 at 12:41
• " the question does not really specify the center of the power series representation. does that mean that my answer is actually correct as well?" In a word, no because when the center is not specified one is supposed to understand the center is zero. (Additionnally, in some curricula the only admissible center is zero.) – Did Aug 16 '15 at 15:59

Hint. Your route is OK, but you should rather start with $$\ln(3-x) = -\int_0^x { \frac{1}{3-t} dt}+\ln 3$$ then follow the same path to obtain the right answer.

• Thanks! I followed this and ended up in the same form. however, say in an exam i wrote like the above, would it be correct though? – Aaron Aug 15 '15 at 12:36

Both series are correct. The one from the lecture is the series expansion around $x=0$, while the one derived in the posted question is the series expansion around $x=2$. And one could choose other arbitrary points around which to expand the function.

Using a straightforward approach we see that for $f(x)=\log(3-x)$, we have for $n>0$

$$f^{(n)}(x)=(-1)^{n+1}(n-1)!(x-3)^{-n} \tag 1$$

We will use this in Approach 2 of the expansions around both $x=0$ and $x=3$ in that which follows.

EXPANSION AROUND $x=0$

Approach 1: Using the approach outlined in the posted question, we find that

\begin{align} \log(3-x)&=-\int_2^x \frac{1}{3-t}dt\\\\ &=-\int_2^x\frac{1}{1-(t-2)}dt\\\\ &=-\sum_{n=0}^{\infty}\int_0^x (t-2)^n\\\\ &=-\sum_{n=1}^{\infty}\frac{(x-2)^n}{n} \end{align}

which converges for $-1\le x<3$ and diverges otherwise.

Approach 2: From $(1)$, we can see that $f^{(n)}(2)=(-1)^{n+1}(n-1)!(-1)^{-n}=-(n-1)!$

Therefore, we can write the series representation as

$$\log(3-x)=-\sum_{n=1}^{\infty}\frac{(x-2)^n}{n}$$

which converges for $-1\le x<3$ and diverges otherwise as expected!

EXPANSION AROUND $x=3$

Approach 1: Using the approach outlined in the posted question, we find that

\begin{align} \log(3-x)&=\log 3-\int_0^x \frac{1}{3-t}dt\\\\ &=\log 3-\frac13\int_0^x\frac{1}{1-(t/3)}dt\\\\ &=\log 3-\frac13\sum_{n=0}^{\infty}\int_0^x (t/3)^n\\\\ &=\log 3-\sum_{n=1}^{\infty}\frac{x^n}{n3^n} \end{align}

which converges for $-3\le x<3$ and diverges otherwise.

Approach 2: From $(1)$, we can also see that $f^{(n)}(0)=(-1)^{n+1}(n-1)!(-3)^{-n}=-\frac{(n-1)!}{3^n}$.

Therefore, we can write the series representation as

$$f(x)=\log 3-\sum_{n=1}^{\infty}\frac{x^n}{n3^n}$$

which converges for $-3\le x<3$ and diverges otherwise as expected!

• Please let me know how I can improve my answer. I really just want to give you the best answer I can. – Mark Viola Aug 17 '15 at 15:10


Given the GCD and LCM of n positive integers, how many solutions are there?

Question: Suppose you know $$G:=\gcd$$ (greatest common divisor) and $$L:=\text{lcm}$$ (least common multiple) of $$n$$ positive integers; how many solution sets exist?

In the case of $$n = 2$$, one finds that for the $$k$$ distinct primes dividing $$L/G$$, there are a total of $$2^{k-1}$$ unique solutions.

I am happy to write out a proof of the $$n = 2$$ case if desirable, but my question here concerns the more general version. The $$n=3$$ case already proved thorny in my explorations, so I would be happy to see smaller cases worked out even if responders are unsure about the full generalization.

Alternatively: If there is already an existing reference to this problem and its solution, then a pointer to such information would be most welcome, too!

• @Yorch Your comment only links to the question in the case where $n=2$; for me, this case was no trouble! I am asking, specifically, about the general case: Where you have positive integers $\{a_1, \ldots, a_n\}$. Sep 22 '21 at 15:53
• do you require that the $n$ positive integers be distinct? Are you trying to count the multisets? I think that is the only version I haven't been able to solve. Sep 22 '21 at 16:04
• @Yorch No requirement that the integers be distinct and/but (ideally!) counting distinct solutions. If you think that you can make traction on a modified version (i.e. imposing additional constraints) then I'd still be pleased to see what you come up with. Sep 22 '21 at 16:08

If you are interested in counting tuples $$(a_1,a_2,\dots,a_n)$$ such that $$\gcd(a_1,\dots,a_n) = G$$ and $$\operatorname{lcm}(a_1,\dots,a_n) = L$$ then we can do it as follows.

If $$L/G = \prod\limits_{i=1}^s p_i^{x_i}$$ then each $$a_i$$ must be of the form $$G \prod\limits_{j=1}^s p_i^{y_{i,j}}$$ with $$0 \leq y_{i,j} \leq x_i$$.

Hence for each prime $$p_i$$ we require that the function from $$\{1,\dots, n\}$$ to $$\mathbb N$$ that sends $$j$$ to $$y_{i,j}$$ be a function that hits $$0$$ and $$x_i$$.

The number of such functions is easy by inclusion-exclusion for $$x_i \geq 1$$, it is $$(x_i+1)^n - 2(x_i)^n + (x_i-1)^n$$.

It follows the total number of tuples is $$\prod\limits_{i=1}^s ( (x_i+1)^n - 2x_i^n + (x_i-1)^n)$$.

• Counting tuples as in, with repetition, right? E.g. $(1,2)$ and $(2,1)$ would each be counted in your computation? If so, isn't it the case that (using your notation) you could assign the $s$ distinct primes (to their various powers) as divisors of any of the $n$ integers or a subset of them (e.g. to $\{a_1, a_3, a_7\}$)? There are $2^n$ subsets of $\{a_1, \ldots, a_n\}$, but we exclude the full set (this is the $\gcd$) as well as the empty set for a total of $2^{n} - 2$ subsets. Assigning the aforementioned $s$ primes can now be done in in $s^{2^{n} - 2}$ ways. Or have I misunderstood? Sep 22 '21 at 17:17
• Yes, that is what it looks like when no prime appears more than once in $L/G$, you would get $(2^n-2)^s$@BenjaminDickman , when you have a prime with exponent greater than $1$ dividing $L/G$ it becomes more complex. Sep 22 '21 at 17:49
• lets consider $G=1$ and $L=8$ and $n = 3$. Here we must have that each $a_i$ is one of $1,2,4,8$, and we require that at least one of them is $1$ and at least one of them is $8$, there are $4^3$ total tuples, there are $3^3$ tuples that don't hit the value one, there are $3^3$ that don't hit the value $8$ and there are $2^3$ that don't hit etiher, so there are $4^3-2(3^3) + 2^3$ total triples that work. Sep 22 '21 at 18:08
• Ah, great! I have also been pointed to this same answer as Theorem 2.7 here: derby.openrepository.com/handle/10545/583372 (I may add an answer to this effect) Sep 22 '21 at 19:34
• The case $G,L$ is the same as the case $1,L/G$ Sep 23 '21 at 16:39

(Adding this community wiki answer to point out a relevant reference.) I was recently pointed to the following paper, in which this and related problems are proposed and solved:

Bagdasar, O. (2014.) "On some functions involving the lcm and gcd of integer tuples." Scientific Publications of the State University of Novi Pazar Series A: Applied Mathematics, Informatics and mechanics, 6(2):91-100. PDF (no paywall).

The result appears as Theorem 2.7 (cf. the comment of Yorch, too):


# Inferential logic in a simple-life situation.

Here's a little situation I want math to resolve for me :

1. If I study, I make the exam ,
2. If I do not play tennis, I study ,
3. I didn't make the exam

Can I conclude that was playing tennis ?

Trying to put this into the symbology of inference logic and propositional classic logic :

$P1 : \text{study} \implies \text{exam}$

$P2 : (\text{tennis}\, \vee \text{study}) \wedge (\neg \text{tennis} \implies \text{study})$ (disjunctive syllogism)

$p3 : \neg \text{exam}$

My reasoning :

Step 1 : the contrapositive of $P1$ is $P1' : \neg \text{exam} \implies \neg \text{study}$ ;

Step 2 : By Modus Tollens ( $[(P \implies Q) \wedge \neg Q] \implies \neg P$) we have : $(\text{study} \implies \text{exam}) \wedge (\neg \text{exam} \implies \neg \text{study})$

Step 3 : should we suppose : $\neg \text{tennis} \wedge \neg \text{study}$, then $\neg ( \text{tennis} \vee \text{study})$, then (by $P2$) $\text{tennis}$ or otherwise the $P1$ would fall since $\neg \text{study}$ and $\neg (False \implies False)$.

Step 4 : reductio ad absurdum from step $(3)$, we have $(\text{tennis} \vee \text{study})$, henceforth, in $P2$, $\neg \text{tennis}$ or else $false \implies false$.

So, have I been playing tennis or is my inferential logic bad ?

• The title should be more informative. – Paracosmiste Dec 25 '13 at 15:38
• Would the person that "minused" the question care to say why ? That would be nice ! – Gloserio Dec 25 '13 at 16:11
• I'm not the downvoter. – Paracosmiste Dec 25 '13 at 16:26
• I am not accusing either, and I've just asked the question to see what I can avoid next time :) – Gloserio Dec 25 '13 at 16:28

Yes, indeed, we can easily arrive at the conclusion that you played tennis: and the repeated use of modus tollens, alone (plus one invocation of double negation) gets you that conclusion.

Our premises, in "natural language":

1. If I study, I make the exam ,
2. If I do not play tennis, I study ,
3. I didn't make the exam

KEY:

$S:\;$ I study.

$E:\;$ I make the exam.

$P:\;$ I play tennis.

Then our premises translate to:

$(1): S \rightarrow E$.

$(2): \lnot P \rightarrow S.$

$(3): \lnot E.$

$(4)\quad \lnot S$ follows from $(1), (3)$ by modus tollens.

$(5)\quad \lnot \lnot P$ follows from $(2), (4)$ by modus tollens.

$\therefore (6) \quad P$, by $(5)$ and double negation.

Hence you can conclude you played tennis.

• As @Matt Brenneman did, you translated the second statement to : $\neg P \implies S$ while I've translated it to : $(P \vee S) \wedge \neg S \implies P$, which I thought was safer. Admitting your translation, I would totally agree with your reasoning, but admitting mine, would we come still to the conclusion that I play tennis ? – Gloserio Dec 25 '13 at 16:19
• Yes, absolutely you would! $\lnot P \rightarrow S \equiv \lnot \lnot P \lor S\equiv P \lor S$. Then since we have $\lnot S$, too, you can conclude $P$. – Namaste Dec 25 '13 at 16:36
• I know you love logic and for this reason this answer is excellent;-)+1 – user63181 Dec 25 '13 at 16:51
• @amWhy : thank you, now it's clear ! – Gloserio Dec 25 '13 at 18:16
• You're welcome, @Gloserio! – Namaste Dec 25 '13 at 18:23

You made a mistake in your step 3, because considering only P2, the term $\neg ( tennis \vee study)$ does not imply $tennis \vee study$.

This may sound counter-intuitive to your introduction.

The reason is, that your P2 is an arguable translation of statement 2. It is not equivalent to "if I don't play tennis, I study". Rather it states "if I don't play tennis and if I study, I study", which is a tautology. You can see this by drawing a truth-table for P2.

Also note that $\neg exam, \neg tennis, \neg study$ satisfies P1,P2 and P3.

So I would replace P2 by $$\neg tennis \rightarrow study.$$ This also repairs your Step3.

• +1, you're probably true, that why I've asked this question, because I felt as if my $P2$ was somewhat redundant. Thanks for pointing it out ! – Gloserio Dec 25 '13 at 16:26

Yes. It just reduces down to look at the contrapositives of your statements.
Statement 1 is logically equivalent to : ~(make exam) implies ~study.
Statement 2 is logically equivalent to: ~study implies (play tennis).
So the truth of ~(make exam) directly implies you played tennis (use modus ponens twice).

• How is second statement logically equivalent to $\neg study \implies tennis$ ? – Gloserio Dec 25 '13 at 15:52
• It is the contrapositive of the statement: "~(play tennis) implies study" – Matt Brenneman Dec 25 '13 at 15:54
• So how that you converted natural langage to logic symobols, and in its valid. – Gloserio Dec 25 '13 at 16:20


# calculating total number of allowable paths

I seem to be struggling with the following type of path questions

Consider paths starting at $$(0, 0)$$ with allowable steps

(i) from $$(x,y)$$ to $$(x+1,y+2)$$,

(ii) from $$(x,y)$$ to $$(x+2,y+1)$$,

(iii)from $$(x,y)$$ to $$(x+1,y)$$

Determine the total number of allowable paths from $$(0, 0)$$ to $$(8, 8)$$, and the total number of allowable paths from $$(0, 0)$$ to $$(10, 10)$$.

could anyone recommend a trivial method to tackle problems of these types in an exam setting?

• General answer for these type of problems would be to use recursion, as answered by Rob Pratt. However, in this "small" case it might be easier to do things "on hand", especially in exam setting. Suppose you do $A$ moves of type (i), $B$ of type (ii) and $C$ of type (iii). After putting constraints on $A, B, C$ you will see that there is only two possibilities in both of your question. Can you work out the rest by yourself? Apr 2, 2020 at 21:56
• I don't understand @prosinac Apr 2, 2020 at 22:44

Draw a table and think backwards. Let $$p(x,y)$$ be the number of such paths from $$(0,0)$$ to $$(x,y)$$. By conditioning on the last step into $$(x,y)$$, we find that $$p(x,y)=p(x-1,y-2)+p(x-2,y-1)+p(x-1,y),$$ where $$p(x,y)=0$$ if $$x<0$$ or $$y<0$$. You know that $$p(0,0)=1$$, and you want to compute $$p(8,8)$$ and $$p(10,10)$$. The resulting table is $$\begin{matrix} x\backslash y &0 &1 &2 &3 &4 &5 &6 &7 &8 &9 &10 \\ \hline 0 &1 &0 &0 &0 &0 &0 &0 &0 &0 &0 &0 \\ 1 &1 &0 &1 &0 &0 &0 &0 &0 &0 &0 &0 \\ 2 &1 &1 &2 &0 &1 &0 &0 &0 &0 &0 &0 \\ 3 &1 &2 &3 &2 &3 &0 &1 &0 &0 &0 &0 \\ 4 &1 &3 &5 &6 &6 &3 &4 &0 &1 &0 &0 \\ 5 &1 & &8 &12 &13 &12 &10 &4 &5 &0 &1 \\ 6 & & &12 & &27 &30 &26 &20 &15 &5 &6 \\ 7 & & & & &51 & &65 &60 &45 &30 &21 \\ 8 & & & & & & &146 & &\color{red}{130} &105 &71 \\ 9 & & & & & & & & &336 & &231 \\ 10 & & & & & & & & & & &\color{red}{672} \\ \end{matrix}$$ In particular, $$p(8,8) = p(7,6)+p(6,7)+p(7,8) = 65+20+45=130.$$

• How could you apply this with perhaps a path D: $(x,y)->(x,y-1)$ ? Apr 3, 2020 at 13:13
• You can use the same approach. The new recurrence has an additional $+p(x,y+1)$ term, and you should explicitly include a boundary condition $p(x,y)=0$ for $y>2x$ to avoid infinite descent. Apr 3, 2020 at 14:46

Suppose you do $$A$$ moves of type (i), $$B$$ of type (ii) and $$C$$ of type (iii). If you want to reach $$(8, 8)$$, then clearly $$A + 2B + C = 8$$ and $$2A + B = 8$$. This yields $$B = 8-2A$$ and $$C = 3A - 8$$. Since $$A, B, C$$ are nonnegative integers, you obtain solutions $$(A, B, C) = (3, 2, 1)$$ or $$(4, 0, 4)$$.

Now to calculate paths, for $$(4, 0, 4)$$ case, a path is described by string of four $$A$$s and four $$C$$s. For example, $$AAACCACC$$ is one such path. There is $${8 \choose 4} = 70$$ such paths.

For the $$(3, 2, 1)$$ case there is $${6 \choose 3} \cdot 3 = 60$$ such paths. Altogether there are $$130$$ such paths.

For reaching $$(10, 10)$$ same logic gives you $$B = 10 - 2A$$ and $$C = 3A - 10$$, so solutions are $$(4, 2, 2)$$ and $$(5, 0, 5)$$.

So, the number of paths is $${8 \choose 4} {4 \choose 2} + {10 \choose 5} = 70 \cdot 6 + 252 = 672$$.

This is in agreement with Rob Pratts answer. Also, I would like to emphasise as I did in the comment that his answer is "better" in a sense that it illustrates how you can handle any problem of this type, since his answers scales reasonably to larger numbers. This answer can from the practical point of view only be used on such a small examples. But, if I was writing an exam, I'd take this approach (or at least I would try and then estimate would it be faster done this way or in the more general way)

• +1 I like these kind of "restrict by case" approaches too. It is not that hard to scale for larger numbers as there is a pattern, though the calculation could be tedious. E.g It works nicely for This question. Apr 3, 2020 at 0:08
• If you prefer multinomial coefficients to binomial coefficients, the formulas are $\binom{6}{3,2,1}+\binom{8}{4,4}$ and $\binom{8}{4,2,2}+\binom{10}{5,5}$. Apr 3, 2020 at 1:24
• I understand everything but where has A+2B+C=8 and 2A+B=8 come from? I don't seen how 8 can be reached using these steps? @prosinac Apr 3, 2020 at 16:58
• If you do $A$ moves of type (i) then you moved $A$ steps on the $x$-axis. If you do $B$ moves of type (ii) then you moved $2B$ steps on the $x$-axis. If you do $C$ moves of type (iii) then you moved $C$ steps on the $x$-axis. And you must move $8$ steps in total. So $A+2B+C=8$. Same logic for $y$-axis gives second equation Apr 3, 2020 at 21:26


It is currently 19 Feb 2018, 04:06

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

# M is a positive integer less than 100. When m is raised to the third

Author Message
TAGS:

### Hide Tags

Manager
Joined: 03 Oct 2016
Posts: 84
Concentration: Technology, General Management
WE: Information Technology (Computer Software)
M is a positive integer less than 100. When m is raised to the third [#permalink]

### Show Tags

06 Oct 2016, 12:40
2
KUDOS
19
This post was
BOOKMARKED
00:00

Difficulty:

65% (hard)

Question Stats:

67% (02:01) correct 33% (02:19) wrong based on 192 sessions

### HideShow timer Statistics

M is a positive integer less than 100. When m is raised to the third power, it becomes the square of another integer. How many different values could m be?

A. 7
B. 9
C. 11
D. 13
E. 15

Keep the Kudos dropping in and let these tricky questions come out ....
[Reveal] Spoiler: OA

_________________

KINDLY KUDOS IF YOU LIKE THE POST

SC Moderator
Joined: 13 Apr 2015
Posts: 1578
Location: India
Concentration: Strategy, General Management
WE: Analyst (Retail)
Re: M is a positive integer less than 100. When m is raised to the third [#permalink]

### Show Tags

06 Oct 2016, 19:13
2
KUDOS
$$({a^2})^3 = a^6 = ({a^3})^2$$

Given: M = $$0 < a^2 < 100$$

Values of a^2 can be --> $$1^2, 2^2, 2^4, 2^6, 3^2, 3^4, 5^2, 7^2, (2^2 * 3^2)$$

There are 9 possible values.

Manager
Joined: 26 Jun 2013
Posts: 92
Location: India
Schools: ISB '19, IIMA , IIMB
GMAT 1: 590 Q42 V29
GPA: 4
WE: Information Technology (Retail Banking)
Re: M is a positive integer less than 100. When m is raised to the third [#permalink]

### Show Tags

01 Mar 2017, 11:35
Vyshak wrote:
$$({a^2})^3 = a^6 = ({a^3})^2$$

Given: M = $$0 < a^2 < 100$$

Values of a^2 can be --> $$1^2, 2^2, 2^4, 2^6, 3^2, 3^4, 5^2, 7^2, (2^2 * 3^2)$$

There are 9 possible values.

Vyshak please can you explain this as I am not able to understand.

Thanks!
_________________

Remember, if it is a GMAT question, it can be simplified elegantly.

SC Moderator
Joined: 13 Apr 2015
Posts: 1578
Location: India
Concentration: Strategy, General Management
WE: Analyst (Retail)
Re: M is a positive integer less than 100. When m is raised to the third [#permalink]

### Show Tags

01 Mar 2017, 22:47
1
KUDOS
hotshot02 wrote:

Vyshak please can you explain this as I am not able to understand.

Thanks!

You have to find the number of values of a^2 that are between 0 and 100. --> a is between 0 and 10 --> We will have 9 values for m.

Hope it helps.
Target Test Prep Representative
Affiliations: Target Test Prep
Joined: 04 Mar 2011
Posts: 1975
Re: M is a positive integer less than 100. When m is raised to the third [#permalink]

### Show Tags

06 Mar 2017, 17:14
1
KUDOS
Expert's post
4
This post was
BOOKMARKED
idontknowwhy94 wrote:
M is a positive integer less than 100. When m is raised to the third power, it becomes the square of another integer.
How many different values could m be?

A. 7
B. 9
C. 11
D. 13
E. 15

We are given that m is a positive integer less than 100. We are also given that when m is raised to the third power, it becomes the square of another integer. In order for that to be true, m itself must (already) be a perfect square, since any perfect square raised to the third power will still be a perfect square, i.e., square of an integer. Thus we are looking for perfect squares that are less than 100. Since there are 9 perfect squares that are less than 100, namely, 1, 4, 9, …, 64, and 81, the answer is 9.

Let’s look at some examples to clarify this: Let’s assume that m = 4 = 2^2. Now, let’s raise m to the third power, obtaining m^3 = (2^2)^3 = 4^3 = 64, which is the perfect square 8^2. Another illustration: Let’s let m = 25 = 5^2. Now, let’s raise m to the third power, obtaining m^3 = (5^2)^3 = 25^3 = 15625, which is the perfect square of 125.

(Note: By the way the problem is worded, “when m is raised to the third power, it becomes the square of another integer,” 1 should not be counted as one of the 9 different values m could be, unlike all the other 8 values. For example, take the number 4: 4^3 = 64 = 8^2, which is the square of another integer, 8. However, 1^3 = 1 = 1^2, which is the square of the same integer. The correct way to word the problem is “when m is raised to the third power, it becomes the square of an integer.”)

_________________

Jeffery Miller

GMAT Quant Self-Study Course
500+ lessons 3000+ practice problems 800+ HD solutions

CR Forum Moderator
Status: The best is yet to come.....
Joined: 10 Mar 2013
Posts: 529
Re: M is a positive integer less than 100. When m is raised to the third [#permalink]

### Show Tags

26 Aug 2017, 00:14
JeffTargetTestPrep wrote:
Any perfect square raised to the third power will still be a perfect square, i.e., square of an integer. Thus we are looking for perfect squares that are less than 100.

Why we need to look for perfect squares that are less than 100? We need the values of $$m<100$$ NOT the $$m^2<100$$.

I know I am wrong, but I don't know why I am wrong.
_________________

Hasan Mahmud

Director
Joined: 18 Aug 2016
Posts: 628
Concentration: Strategy, Technology
GMAT 1: 630 Q47 V29
GMAT 2: 740 Q51 V38
Re: M is a positive integer less than 100. When m is raised to the third [#permalink]

### Show Tags

26 Aug 2017, 01:09
idontknowwhy94 wrote:
M is a positive integer less than 100. When m is raised to the third power, it becomes the square of another integer.
How many different values could m be?

A. 7
B. 9
C. 11
D. 13
E. 15

Keep the Kudos dropping in and let these tricky questions come out ....

0<x^2<100
1,4,9,16,25,36,49,64,81
9 numbers
B

Sent from my iPhone using GMAT Club Forum mobile app
_________________

We must try to achieve the best within us

Thanks
Luckisnoexcuse

Senior Manager
Joined: 29 Jun 2017
Posts: 495
GMAT 1: 570 Q49 V19
GPA: 4
WE: Engineering (Transportation)
Re: M is a positive integer less than 100. When m is raised to the third [#permalink]

### Show Tags

10 Sep 2017, 06:38
2
KUDOS
Ans is B

M^3 = y^2
M= y^(2/3)
means y should have a cube-root as integer and whose square should be less than 100
lets say start by 1000 =y
y^(2/3) = 1000^2/3 =100 rejected means M cant be 100
M can attain 81,64,49,36,25,16,09,04,01 TOTAL 9 values .

_________________

Give Kudos for correct answer and/or if you like the solution.

Senior Manager
Status: Countdown Begins...
Joined: 03 Jul 2016
Posts: 308
Location: India
Concentration: Technology, Strategy
Schools: IIMB
GMAT 1: 580 Q48 V22
GPA: 3.7
WE: Information Technology (Consulting)
Re: M is a positive integer less than 100. When m is raised to the third [#permalink]

### Show Tags

21 Oct 2017, 21:01
idontknowwhy94 wrote:
M is a positive integer less than 100. When m is raised to the third power, it becomes the square of another integer.
How many different values could m be?

A. 7
B. 9
C. 11
D. 13
E. 15

Keep the Kudos dropping in and let these tricky questions come out ....

Bunuel,
For this question, I understood that basically we need to find out squares of integers below 100. But I have one query -

Why does above answers includes $$1^2$$? The question stem says if the number is raised to third power, then it becomes square of ANOTHER integer?

$$\sqrt{(1^3)}$$ = 1 only.

_________________

Need Kudos to unlock GMAT Club tests

Intern
Joined: 21 May 2017
Posts: 25
Re: M is a positive integer less than 100. When m is raised to the third [#permalink]

### Show Tags

21 Oct 2017, 22:09
RMD007 wrote:
idontknowwhy94 wrote:
M is a positive integer less than 100. When m is raised to the third power, it becomes the square of another integer.
How many different values could m be?

A. 7
B. 9
C. 11
D. 13
E. 15

Keep the Kudos dropping in and let these tricky questions come out ....

Bunuel,
For this question, I understood that basically we need to find out squares of integers below 100. But I have one query -

Why does above answers includes $$1^2$$? The question stem says if the number is raised to third power, then it becomes square of ANOTHER integer?

$$\sqrt{(1^3)}$$ = 1 only.

JeffTargetTestPrep addressed this in his response above -

(Note: By the way the problem is worded, “when m is raised to the third power, it becomes the square of another integer,” 1 should not be counted as one of the 9 different values m could be, unlike all the other 8 values. For example, take the number 4: 4^3 = 64 = 8^2, which is the square of another integer, 8. However, 1^3 = 1 = 1^2, which is the square of the same integer. The correct way to word the problem is “when m is raised to the third power, it becomes the square of an integer.”)
Senior Manager
Status: Countdown Begins...
Joined: 03 Jul 2016
Posts: 308
Location: India
Concentration: Technology, Strategy
Schools: IIMB
GMAT 1: 580 Q48 V22
GPA: 3.7
WE: Information Technology (Consulting)
Re: M is a positive integer less than 100. When m is raised to the third [#permalink]

### Show Tags

21 Oct 2017, 22:13
mahu101 wrote:

JeffTargetTestPrep addressed this in his response above -

(Note: By the way the problem is worded, “when m is raised to the third power, it becomes the square of another integer,” 1 should not be counted as one of the 9 different values m could be, unlike all the other 8 values. For example, take the number 4: 4^3 = 64 = 8^2, which is the square of another integer, 8. However, 1^3 = 1 = 1^2, which is the square of the same integer. The correct way to word the problem is “when m is raised to the third power, it becomes the square of an integer.”)

Thanks, my point is, with the given question stem 8 should be the answer!
_________________

Need Kudos to unlock GMAT Club tests

Intern
Joined: 21 May 2017
Posts: 25
Re: M is a positive integer less than 100. When m is raised to the third [#permalink]

### Show Tags

21 Oct 2017, 22:19
1
KUDOS
RMD007 wrote:
mahu101 wrote:

JeffTargetTestPrep addressed this in his response above -

(Note: By the way the problem is worded, “when m is raised to the third power, it becomes the square of another integer,” 1 should not be counted as one of the 9 different values m could be, unlike all the other 8 values. For example, take the number 4: 4^3 = 64 = 8^2, which is the square of another integer, 8. However, 1^3 = 1 = 1^2, which is the square of the same integer. The correct way to word the problem is “when m is raised to the third power, it becomes the square of an integer.”)

Thanks, my point is, with the given question stem 8 should be the answer!

Yes, you are right.
Re: M is a positive integer less than 100. When m is raised to the third   [#permalink] 21 Oct 2017, 22:19
Display posts from previous: Sort by


It is currently 18 Oct 2017, 22:56

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

# If p, x, and y are positive integers, y is odd, and p = x^2

Author Message
TAGS:

### Hide Tags

Senior Manager
Joined: 22 Sep 2005
Posts: 278

Kudos [?]: 241 [17], given: 1

If p, x, and y are positive integers, y is odd, and p = x^2 [#permalink]

### Show Tags

14 Aug 2009, 12:49
17
KUDOS
88
This post was
BOOKMARKED
00:00

Difficulty:

95% (hard)

Question Stats:

40% (02:06) correct 60% (02:17) wrong based on 1935 sessions

### HideShow timer Statistics

If p, x, and y are positive integers, y is odd, and p = x^2 + y^2, is x divisible by 4?

(1) When p is divided by 8, the remainder is 5.
(2) x – y = 3
[Reveal] Spoiler: OA

Kudos [?]: 241 [17], given: 1

Manager
Joined: 25 Jul 2009
Posts: 115

Kudos [?]: 262 [42], given: 17

Schools: NYU, NUS, ISB, DUKE, ROSS, DARDEN
Re: PS: Divisible by 4 [#permalink]

### Show Tags

14 Aug 2009, 13:46
42
KUDOS
27
This post was
BOOKMARKED
netcaesar wrote:
If p, x, and y are positive integers, y is odd, and p = x^2 + y^2, is x divisible by 4?

(1) When p is divided by 8, the remainder is 5.
(2) x – y = 3

SOL:

St1:
Here we will have to use a peculiar property of number 8. The square of any odd number when divided by 8 will always yield a remainder of 1!!

This means that y^2 MOD 8 = 1 for all y
=> p MOD 8 = (x^2 + 1) MOD 8 = 5
=> x^2 MOD 8 = 4

Now if x is divisible by 4 then x^2 MOD 8 will be zero. And also x cannot be an odd number as in that case x^2 MOD 8 would become 1. Hence we conclude that x is an even number but also a non-multiple of 4.
=> SUFFICIENT

St2:
x - y = 3
Since y can be any odd number, x could also be either a multiple or a non-multiple of 4.
=> NOT SUFFICIENT

ANS: A
_________________

KUDOS me if I deserve it !!

My GMAT Debrief - 740 (Q50, V39) | My Test-Taking Strategies for GMAT | Sameer's SC Notes

Kudos [?]: 262 [42], given: 17

Math Expert
Joined: 02 Sep 2009
Posts: 41890

Kudos [?]: 128792 [31], given: 12183

Re: PS: Divisible by 4 [#permalink]

### Show Tags

16 Dec 2010, 07:39
31
KUDOS
Expert's post
21
This post was
BOOKMARKED
nonameee wrote:
Can I ask someone to look at this question a provide a solution that doesn't depend on knowing peculiar properties of number 8 or induction?

Thank you.

If p, x, and y are positive integers, y is odd, and p = x^2 + y^2, is x divisible by 4?

(1) When p is divided by 8, the remainder is 5 --> $$p=8q+5=x^2+y^2$$ --> as given that $$y=odd=2k+1$$ --> $$8q+5=x^2+(2k+1)^2$$ --> $$x^2=8q+4-4k^2-4k=4(2q+1-k^2-k)$$.

So, $$x^2=4(2q+1-k^2-k)$$. Now, if $$k=odd$$ then $$2q+1-k^2-k=even+odd-odd-odd=odd$$ and if $$k=even$$ then $$2q+1-k^2-k=even+odd-even-even=odd$$, so in any case $$2q+1-k^2-k=odd$$ --> $$x^2=4*odd$$ --> in order $$x$$ to be multiple of 4 $$x^2$$ must be multiple of 16 but as we see it's not, so $$x$$ is not multiple of 4. Sufficient.

(2) x – y = 3 --> $$x-odd=3$$ --> $$x=even$$ but not sufficient to say whether it's multiple of 4.

_________________

Kudos [?]: 128792 [31], given: 12183

Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 7674

Kudos [?]: 17354 [26], given: 232

Location: Pune, India
Re: PS: Divisible by 4 [#permalink]

### Show Tags

19 Dec 2010, 07:49
26
KUDOS
Expert's post
17
This post was
BOOKMARKED
netcaesar wrote:
If p, x, and y are positive integers, y is odd, and p = x^2 + y^2, is x divisible by 4?

(1) When p is divided by 8, the remainder is 5.
(2) x – y = 3

Such questions can be easily solved keeping the concept of divisibility in mind. Divisibility is nothing but grouping. Lets say if we need to divide 10 by 2, out of 10 marbles, we make groups of 2 marbles each. We can make 5 such groups and nothing will be left over. So quotient is 5 and remainder is 0. Similarly if you divide 11 by 2, you make 5 groups of 2 marbles each and 1 marble is left over. So 5 is quotient and 1 is remainder. For more on these concepts, check out: http://gmatquant.blogspot.com/2010/11/divisibility-and-remainders-if-you.html

First thing that comes to mind is if y is odd, $$y^2$$ is also odd.
If $$y = 2k+1, y^2 = (2k + 1)^2 = 4k^2 + 4k + 1 = 4k(k+1) + 1$$
Since one of k and (k+1) will definitely be even (out of any two consecutive integers, one is always even, the other is always odd), 4k(k+1) will be divisible by 8. So when y^2 is divided by 8, it will leave a 1.

Stmnt 1: When p is divided by 8, the remainder is 5.
When y^2 is divided by 8, remainder is 1. To get a remainder of 5, when x^2 is divided by 8, we should get a remainder of 4.
$$x^2 = 8a + 4$$ (i.e. we can make 'a' groups of 8 and 4 will be leftover)
$$x^2 = 4(2a+1)$$ This implies $$x = 2*\sqrt{Odd Number}$$because (2a+1) is an odd number. Square root of an odd number will also be odd.
Therefore, we can say that x is not divisible by 4. Sufficient.

Stmnt 2: x - y = 3
Since y is odd, we can say that x will be even (Even - Odd = Odd). But whether x is divisible by 2 only or by 4 as well, we cannot say since here we have no constraints on p. Not sufficient.

_________________

Karishma
Veritas Prep | GMAT Instructor
My Blog

Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Kudos [?]: 17354 [26], given: 232 Intern Joined: 30 May 2013 Posts: 4 Kudos [?]: 8 [8], given: 8 Re: If p, x, and y are positive integers, y is odd, and p = x^2 [#permalink] ### Show Tags 15 Sep 2013, 22:13 8 This post received KUDOS I did this question this way. I found it simple. 1. p=x^2+y^2 y is odd p div 8 gives remainder 5. A number which gives remainder 5 when divided by 8 is odd. so (x^2 + y^2)/8 = oddnumber (x^2 + y^2) = 8 * oddnumber (this is an even number without doubt) x^2 + y^2 is even. Since y is odd to get x^2+y^2 even x must also be odd. X is an odd number not divisible by 4 Option A: 1 alone is sufficient Kudos [?]: 8 [8], given: 8 Intern Joined: 02 Sep 2010 Posts: 45 Kudos [?]: 146 [4], given: 17 Location: India Re: PS: Divisible by 4 [#permalink] ### Show Tags 18 Dec 2010, 11:23 4 This post received KUDOS 1 This post was BOOKMARKED maliyeci wrote: Very good solution I did not know this property of 8. Kudos to you. By and induction. 1^2=1 mod 8 say n^2=1 mod 8 (n is an odd number) than if (n+2)^2=1 mod 8 ? (n+2 is the next odd number) (n+2)^2=n^2+4n+4= 1 + 4n + 4 mod 8 4n+4=0 mod 8 because n is an odd number and 4n=4 mod 8. So induction works. So for any odd number n, n^2=1 mod 8 Its not something one shall already know before attacking a question, you may realize properties like this when u start solving a question. Even I didn't know about this property of 8. I approached the question in following way: Stmt 1: P/8=(x^2+y^2)/8; using remainder theorem; rem[(x^2+y^2)/8]= rem[x^2/8] + rem[y^2/8] if x is divisible by 4, then x^2= 4k*4k= 16K=8*2K is also divisible by 8. now to anaylze rem[y^2/8]; start putting suitable values of y; i.e all odd values starting from 1. for y=1; rem(1/8)=1 for y=3; rem(9/8)=1 for y=5;rem(25/8)=1 so you observe this pattern here. coming back to ques now, as rem[(x^2+y^2)/8]= rem[x^2/8] + rem[y^2/8]= rem[x^2/8] + 1 =5; this means rem[x^2/8] is not 0; which implies x is not divisible my 8; Sufficient Stmt2: y being odd can be accept both 3 and 5 as values and we get different results; thus Insufficient Thus OA is A _________________ The world ain't all sunshine and rainbows. It's a very mean and nasty place and I don't care how tough you are it will beat you to your knees and keep you there permanently if you let it. You, me, or nobody is gonna hit as hard as life. But it ain't about how hard ya hit. It's about how hard you can get it and keep moving forward. How much you can take and keep moving forward. That's how winning is done! Kudos [?]: 146 [4], given: 17 Intern Joined: 14 May 2013 Posts: 12 Kudos [?]: 18 [2], given: 3 Re: If p, x, and y are positive integers, y is odd, and p = x^2 [#permalink] ### Show Tags 12 Jun 2013, 10:58 2 This post received KUDOS netcaesar wrote: If p, x, and y are positive integers, y is odd, and p = x^2 + y^2, is x divisible by 4? (1) When p is divided by 8, the remainder is 5. (2) x – y = 3 1. As p = 8I + 5 we have values of P = 5,13,21,29..... etc .. as y is odd when we solve this p(odd) = x^2 + y^2(odd) x^2 = odd -odd = even which can be 2,4,6 ... etc but if we check for any value of p we don't get any multiple of 4. so it say's clearly that x is not divisible by 4. 2. x-y = 3 x = y(odd)+3 x is even which can be 2,4,6.. so it's not sufficient .. Ans : A _________________ Chauahan Gaurav Keep Smiling Kudos [?]: 18 [2], given: 3 Intern Joined: 25 Jun 2013 Posts: 6 Kudos [?]: 3 [2], given: 0 Re: If p, x, and y are positive integers, y is odd, and p = x^2 [#permalink] ### Show Tags 13 Sep 2013, 20:25 2 This post received KUDOS 1 This post was BOOKMARKED from first statement p = 8j + 5 Put j as 1, 2,3,4,5... p would be 13, 21,29, 37,45... Now in the formula p= x^2+y^2 put 1,3,5,7 as value of y ( as y is odd) to get x. You will notic the possible value of x is 2 which is not divisble by 4. Posted from GMAT ToolKit Kudos [?]: 3 [2], given: 0 Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 7674 Kudos [?]: 17354 [2], given: 232 Location: Pune, India Re: If p, x, and y are positive integers, y is odd, and p = x^2 [#permalink] ### Show Tags 18 Aug 2014, 02:43 2 This post received KUDOS Expert's post 1 This post was BOOKMARKED alphonsa wrote: For statement 1 , wouldn't plugging in values be a better option? No. When you need to establish something, plugging in values is not fool proof. Anyway, in this question, how will you plug in values? You cannot assume a value for x since that is what you need to find. You will assume a value for y and a value for p such that they satisfy all conditions. This itself will be quite tricky. Then when you do get a value for x, you will find that it will be even but not divisible by 4. How can you be sure that this will hold for every value of y and p? When a statement is not sufficient, plugging in values can work - you find two opposite cases - one which answers in yes and the other which answers in no. Then you know that the statement alone is not sufficient. But when the statement is sufficient, it is very hard to prove that it will hold for all possible values using number plugging alone. You need to use logic in that case. _________________ Karishma Veritas Prep | GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for$199

Veritas Prep Reviews

Kudos [?]: 17354 [2], given: 232

Senior Manager
Joined: 23 Jun 2009
Posts: 360

Kudos [?]: 134 [1], given: 80

Location: Turkey
Schools: UPenn, UMich, HKS, UCB, Chicago
Re: PS: Divisible by 4 [#permalink]

### Show Tags

15 Aug 2009, 13:49
1
KUDOS
Very good solution I did not know this property of 8. Kudos to you.

By and induction.
1^2=1 mod 8
say
n^2=1 mod 8 (n is an odd number)
than
if (n+2)^2=1 mod 8 ? (n+2 is the next odd number)
(n+2)^2=n^2+4n+4= 1 + 4n + 4 mod 8
4n+4=0 mod 8 because n is an odd number and 4n=4 mod 8.
So induction works.

So for any odd number n, n^2=1 mod 8

Kudos [?]: 134 [1], given: 80

Director
Joined: 23 Apr 2010
Posts: 573

Kudos [?]: 95 [1], given: 7

Re: PS: Divisible by 4 [#permalink]

### Show Tags

16 Dec 2010, 07:13
1
KUDOS
Can I ask someone to look at this question a provide a solution that doesn't depend on knowing peculiar properties of number 8 or induction?

Thank you.

Kudos [?]: 95 [1], given: 7

Intern
Joined: 25 Mar 2012
Posts: 3

Kudos [?]: 3 [1], given: 1

Re: PS: Divisible by 4 [#permalink]

### Show Tags

16 Jul 2012, 18:19
1
KUDOS
Am i missing something, why cant we take stmt 2 as follows:
squaring x-y=3 on both sides, we get p=9+2xy, that is p=odd + even = odd, not divisible by 4

Kudos [?]: 3 [1], given: 1

Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 7674

Kudos [?]: 17354 [1], given: 232

Location: Pune, India
Re: If p, x, and y are positive integers, y is odd, and p = x^2 [#permalink]

### Show Tags

03 May 2017, 05:31
1
KUDOS
Expert's post
VeritasPrepKarishma wrote:
aliasjit wrote:
I am a little confused about solving the problem:

Stmnt 2: x-y =3

we know y is odd.
and if Y is odd as per problem statement y cannot be anything but 1
as x and y both are positive integers.

Therefore x =4 and is divisible by 4.

y is odd, yes, but why must y be 1?

Responding to a pm:
Quote:
because as per statement 2 if x-y =3 and x and y are both +ve integers could y be anything but 1

But as per statement 2, we do not know that x must be 4. x must be even since y is odd and difference between x and y is odd. But will it be 4, we do not know.
_________________

Karishma
Veritas Prep | GMAT Instructor
My Blog

Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Kudos [?]: 17354 [1], given: 232 Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 7674 Kudos [?]: 17354 [0], given: 232 Location: Pune, India Re: PS: Divisible by 4 [#permalink] ### Show Tags 16 Jul 2012, 23:24 Eshaninan wrote: Am i missing something, why cant we take stmt 2 as follows: squaring x-y=3 on both sides, we get p=9+2xy, that is p=odd + even = odd, not divisible by 4 The question is: "Is x divisible by 4?" not "Is p divisible by 4?" x is even since y is odd. We don't know whether x is divisible by only 2 or 4 as well. _________________ Karishma Veritas Prep | GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for$199

Veritas Prep Reviews

Kudos [?]: 17354 [0], given: 232

Manager
Joined: 29 Jun 2011
Posts: 159

Kudos [?]: 24 [0], given: 29

WE 1: Information Technology(Retail)
Re: If p, x, and y are positive integers, y is odd, and p = x^2 [#permalink]

### Show Tags

03 Sep 2013, 04:00
Excellent explanation Bunuel & Karishma:):)

Kudos [?]: 24 [0], given: 29

Intern
Joined: 21 Sep 2013
Posts: 9

Kudos [?]: 2 [0], given: 0

Re: If p, x, and y are positive integers, y is odd, and p = x^2 [#permalink]

### Show Tags

23 Dec 2013, 23:45
For Statement 1:
since p when divided by 8 leaves remainder 5.We obtain the following equation
p= 8q+5
We know y is odd. If we write p =x^2+y^2 then we get the eqn:
x^2+y^2=8q+5
Since, y is odd, 8q is even and 5 is odd. We get 8q+5 is odd.
Then x^2= odd - y^2
i.e x^2=even
ie x= even
But it's not sufficient to answer the question whether x is a multiple of 4?
By this logic i get E as my answer.
Statement 2: is insufficient.

Kudos [?]: 2 [0], given: 0

Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 7674

Kudos [?]: 17354 [0], given: 232

Location: Pune, India
Re: If p, x, and y are positive integers, y is odd, and p = x^2 [#permalink]

### Show Tags

30 Dec 2013, 23:39
Expert's post
1
This post was
BOOKMARKED
Abheek wrote:
For Statement 1:
since p when divided by 8 leaves remainder 5.We obtain the following equation
p= 8q+5
We know y is odd. If we write p =x^2+y^2 then we get the eqn:
x^2+y^2=8q+5
Since, y is odd, 8q is even and 5 is odd. We get 8q+5 is odd.
Then x^2= odd - y^2
i.e x^2=even
ie x= even
But it's not sufficient to answer the question whether x is a multiple of 4?

Your analysis till now is fine but it is incomplete. We do get that x is even but we also get that x is a multiple of 2 but not 4 as explained in the post above: if-p-x-and-y-are-positive-integers-y-is-odd-and-p-x-82399.html#p837890
_________________

Karishma
Veritas Prep | GMAT Instructor
My Blog

Get started with Veritas Prep GMAT On Demand for \$199

Veritas Prep Reviews

Kudos [?]: 17354 [0], given: 232

Manager
Joined: 22 Jul 2014
Posts: 130

Kudos [?]: 297 [0], given: 197

Concentration: General Management, Finance
GMAT 1: 670 Q48 V34
WE: Engineering (Energy and Utilities)
Re: If p, x, and y are positive integers, y is odd, and p = x^2 [#permalink]

### Show Tags

16 Aug 2014, 00:16
For statement 1 , wouldn't plugging in values be a better option?

Kudos [?]: 297 [0], given: 197

Current Student
Joined: 17 Jul 2013
Posts: 48

Kudos [?]: 33 [0], given: 19

GMAT 1: 710 Q49 V38
GRE 1: 326 Q166 V160
GPA: 3.74
Re: If p, x, and y are positive integers, y is odd, and p = x^2 [#permalink]

### Show Tags

29 Aug 2014, 05:25
Hi Karishma,

Thanks for the explanation to the question. I was just wondering how the answer would change if we change the question stem a little bit. What if the question asks if p (instead of x) is divisible by 4?

In this scenario, statement 1 would be sufficient since if something leaves a remainder of 5, it would leave a remainder of 1 upon division by 4

For statement 2, we know that x = y+3, so x is even. If we square it, it would surely be divisible by 4. Now if a number (y^2, which is odd) non-divisible by 4 is added to a number divisible by 4, the result would surely be not divisible by 4. So statement 2 would also be sufficient.

Is this reasoning correct? just for practicing the concept

Kudos [?]: 33 [0], given: 19

Manager
Joined: 22 Jan 2014
Posts: 141

Kudos [?]: 74 [0], given: 145

WE: Project Management (Computer Hardware)
Re: If p, x, and y are positive integers, y is odd, and p = x^2 [#permalink]

### Show Tags

31 Aug 2014, 03:38
netcaesar wrote:
If p, x, and y are positive integers, y is odd, and p = x^2 + y^2, is x divisible by 4?

(1) When p is divided by 8, the remainder is 5.
(2) x – y = 3

1) p = 8k+5 (k is a whole number)
also p = (x^2+y^2)
=> (x^2+y^2) mod 8 = 5

any square (n^2) mod 8 follows the following pattern -> 1,4,1,0 and then repeats.
for getting x^2+y^2 mod 8 = 5
we need to take a 4 and a 1 from the above pattern. at multiples of 4, the remainder is 0. so x can never be divisible by 4.
A is sufficient.

2) x-y=3 (odd)
even - odd = odd
or odd - even = odd
so insufficient.

Hence, A.
_________________

Illegitimi non carborundum.

Kudos [?]: 74 [0], given: 145

Re: If p, x, and y are positive integers, y is odd, and p = x^2   [#permalink] 31 Aug 2014, 03:38

Go to page    1   2    Next  [ 32 posts ]

Display posts from previous: Sort by


# What does "curly (curved) less than" sign $\succcurlyeq$ mean?

I am reading Boyd & Vandenberghe's Convex Optimization. The authors use curved greater than or equal to (\succcurlyeq)

$$f(x^*) \succcurlyeq \alpha$$

and curved less than or equal to (\preccurlyeq)

$$f(x^*) \preccurlyeq \alpha$$

Can someone explain what they mean?

• as () is curved and {} are curly ( en.wikipedia.org/wiki/Bracket ), I think those symbols you mention are curved not curly Feb 9, 2014 at 6:43
• @barlop If you look at $\LaTeX$ source of the formulas in question (right click->Show Math As->TeX commands), you'll see \succcurlyeq, which has curly word in it, not curved. Feb 9, 2014 at 7:18
• Thanks everyone for answers. As I understand $\succeq$ or $\preceq$ are more general than their more popular counterparts. I think Michael's answer make sense. If I understand correctly, $X \succeq Y, \quad if \quad \| X \| \ge \| Y \|$ where $\| \cdot \|$ is the norm associated with the space $X$ and $Y$ belongs to. I think Chris's answer is correct but is more strict condition than Michael's answer. Please correct me if I'm wrong. Feb 9, 2014 at 17:53
• You should definitely take the tour. This is not a traditional forum!
– bodo
Feb 9, 2014 at 18:03
• Well I don't know what a traditional forum looks like. :-) Feb 9, 2014 at 18:20

Both Chris Culter's and Code Guru's answers are good, and I've voted them both up. I hope that I'm not being inappropriate by combining and expanding upon them here.

It should be noted that the book does not use $\succeq$, $\preceq$, $\succ$, and $\prec$ with scalar inequalities; for these, good old-fashioned inequality symbols suffice. It is only when the quantities on the left- and right-hand sides are vectors, matrices, or other multi-dimensional objects that this notation is called for.

The book refers to these relations as generalized inequalities, but as Code-Guru rightly points out, they have been in use for some time to represent partial orderings. And indeed, that's exactly what they are, and the book does refer to them that way as well. But given that the text deals with convex optimization, it was apparently considered helpful to refer to them as inequalities.

Let $S$ be a vector space, and let $K\subset S$ be a closed, convex, and pointed cone with a non-empty interior. (By cone, we mean that $\alpha K\equiv K$ for all $\alpha>0$; and by pointed, we mean that $K\cap-K=\{0\}$.) Such a cone $K$ induces a partial ordering on the set $S$, and an associated set of generalized inequalities: $$x \succeq_K y \quad\Longleftrightarrow\quad y \preceq_K x \quad\Longleftrightarrow\quad x - y \in K$$ $$x \succ_K y \quad\Longleftrightarrow\quad y \prec_K x \quad\Longleftrightarrow\quad x - y \in \mathop{\textrm{Int}} K$$ This is a partial ordering because, for many pairs $x,y\in S$, $x \not\succeq_K y$ and $y \not\succeq_K x$. So that's the primary reason why he and others prefer to use the curly inequalities to denote these orderings, reserving $\geq$, $\leq$, etc. for total orderings. But it has many of the properties of a standard inequality, such as: $$x\succeq_K y \quad\Longrightarrow\quad \alpha x \succeq_K \alpha y \quad\forall \alpha>0$$ $$x\succeq_K y \quad\Longrightarrow\quad \alpha x \preceq_K \alpha y \quad\forall \alpha<0$$ $$x\succeq_K y, ~ x\preceq_K y \quad\Longrightarrow\quad x=y$$ $$x\succ_K y \quad\Longrightarrow\quad x\not\prec_K y$$

When the cone $K$ is understood from context, it is often dropped, leaving only the inequality symbol $\succeq$. There are two cases where this is almost always done. First, when $S=\mathbb{R}^n$ and the cone $K$ is non-negative orthant $\mathbb{R}^n_+$ the generalized inequality is simply an elementwise inequality: $$x \succeq_{\mathbb{R}^n_+} y \quad\Longleftrightarrow\quad x_i\geq y_i,~i=1,2,\dots,n$$ Second, when $S$ is the set of symmetric $n\times n$ matrices and $K$ is the cone of positive semidefinite matrices $\mathcal{S}^n_+=\{X\in S\,|\,\lambda_{\text{min}}(X)\geq 0\}$, the inequality is a linear matrix inequality (LMI): $$X \succeq_{\mathcal{S}^n_+} Y \quad\Longleftrightarrow\quad \lambda_{\text{min}}(X-Y)\geq 0$$ In both of these cases, the cone subscript is almost always dropped.

Many texts in convex optimization don't bother with this distinction, and use $\geq$ and $\leq$ even for LMIs and other partial orderings. I prefer to use it whenever I can, because I think it helps people realize that this is not a standard inequality with an underlying total order. That said, I don't feel that strongly about it for $\mathbb{R}^n_+$; I think most people rightly assume that $x\geq y$ is considered elementwise when $x,y$ are vectors.

• Thanks a lot for the detailed answer, and for correcting the symbol as well :-). Feb 9, 2014 at 16:35
• This is an old answer and I completely agree with it but I thought that providing another common application of this notation might be useful. As @Code-Guru pointed out, these are useful for partial orders. In Economics, we usually model preferences over baskets of goods with "not worse than" or "not better than" sets. The partial order concept fits like a glove in this situation. The interested reader might find books in Decision Sciences useful for this kind of discussion. Dec 30, 2020 at 21:13

There's a list of notation in the back of the book. On page 698, $x\preceq y$ is defined as componentwise inequality between vectors $x$ and $y$. This means that $x_i\leq y_i$ for every index $i$.

Edit: The notation is introduced on page 32.

Often these symbols represent partial order relations. The typical "less than" and "greater than" operations both define partial orders on the real numbers. However, there are many other examples of partial orders.

Sometimes the curly greater than sign is used to indicate positive semi-definiteness of a matrix $$X$$:

$$(X\succeq 0\ \text{or}\ X\ge 0)$$

or a function $$f(x)$$

$$(f(x) \succeq 0\ \text{or}\ f(x)\ge 0).$$

• Positive definiteness of what? Aug 26, 2015 at 17:03

Does it sometimes denote, in measurement theory; often the qualitative counterpart to $\geq$ in the numerical representation; when one wants to numerically represent a totally ordered qualitative probability representation:

$$A ≽ B \leftrightarrow A > B \leftrightarrow F(A) \geq F(B)$$.

On the other hand. I have seen it used in multi-dimensional partial or even total, orderings under the "ordering of major-ization" for vector valued functions or for a system for two kinds of orderings.

One for (numerical) ordinal, $>$ comparisons and another, $≽$ for (numerical) differences, sums, or a some other kind of relation, to fine grain, the representation to ensure (or some kind of) unique-ness, rather than merely strong represent-ability.

See Marshall

Marshall, Albert W.; Olkin, Ingram, Inequalities: theory of majorization and its applications, Mathematics in Science and Engineering, Vol. 143. New York etc.: Academic Press. XX, 569 p. \$52.50 (1979). ZBL0437.26007. Such functions or representations, may use both the curly greater$≽$than and$\geq $in the numerical or functional representation and are compatible with total orderings. $$a,b\in \Omega^{n}\; a <b \,,\quad \text{or}, \quad, a= b, \quad \text{or}\quad a < b$$ where one cannot usually adding up distinct $$a_1 \in \Omega_i\,; b_j\in \Omega_j; j \neq i$$ $$a_3 \in \Omega_c;\, ; \, P(a_i) + P(b_j) = P(a_3)$$. Sometimes I "(conjecture)" that$≽$denotes a weaker version of$(A)$or$(A_1)$(strong complementary add-itivity) which orders the differences and sums, is something over and above$(B)$;strict monotone increasing/strong representability/order embedding- above reflecting and preserving . $$(A)\,[f(x_1)+f(x_2)≽ f(y_1) + f(y_2)] \rightarrow [f(x) \geq f(y)] \text{where} \,; x=x_1 +x_2\,, y=y_1+y_2$$ or $$(A1)\,[x\, \geq y ]\,\rightarrow [f(x_1)+f(x_2)≽ f(y_1) + f(y_2)]\text{where} \,; x=x_1 +x_2\,, y=y_1+y_2$$ Rather than:$$(B)$$ $$(B)\,x\, \geq y \leftrightarrow f(x) \geq f(y)\, \text{where} \,; x=x_1 +x_2\,, y=y_1+y_2$$ In contrast to$(B)$a standard order embedding (strict monotone increasing function) Where in$(B)$the numerical function,$F$or representation is 'merely strong'. That is$F$is a monotone strictly increasing function of some entity$x$where$F(x)$is the entity one wishes to order by$x$. Sometimes, even in a infinite and uniformly and non-atomic, continuous total order representation, nothing unique will come out. If its infinite /non-atomic in the wrong sense. ie in a super-atomic or entangled lexicographic system. Where the system is dense/or continuum dense/non-atomic/bottom-less in the wrong sense. Such as a spin$1/2$system in quantum mechanics, or a lexicographic entangled, system where its infinite in wrong sense, in the vertical, not in the horizontal, not within a basis of space/basis.$(A)$is arguably a lot stronger than$(B)$, or at least is, if relatively unrestricted. Where$(A)$, may be used not to extend the ordering so much as much as make a total ordering 'numerically precise; to put some constraints on sums and differences, where the events are the not kinds of things that usually add up. Say in an entangled multidimensional quantum spin 1/2 system or utility representation where the events are on complementary spaces and mixtures are not allowed, for example. That is, to, put a metric on differences (say on a two outcome system). That is something stronger, than a, total or non atomic/dense order, where(merely) every event. Even if the entire system is totally ordered within and betwixt the distinct$\Omega\$. So that the system is solv-able, and uniquely so.


# Math Help - Piecewise-Defined Function

1. ## Piecewise-Defined Function

For both questions below:

(a) Find the domain of the function.

(b) Locate any intercepts.

(1)

.....{3 + x......if -3 <or= to x < 0
f(x){3...........if......x = 0
.....{Sqrt{x}..if......x > 0

=======================

(2)

..........{1/x.........if.....x < 0
f(x) = {sqrt{x}...if.....x >or= to 0

NOTE: It is hard to correctly type the piecewise-defined functions using a regular keyboard.

I hope you can understand the above.

2. Originally Posted by symmetry
For both questions below:

(a) Find the domain of the function.

(b) Locate any intercepts.

(1)

.....{3 + x......if -3 <or= to x < 0
f(x){3...........if......x = 0
.....{Sqrt{x}..if......x > 0

=======================

(2)

..........{1/x.........if.....x < 0
f(x) = {sqrt{x}...if.....x >or= to 0

NOTE: It is hard to correctly type the piecewise-defined functions using a regular keyboard.

I hope you can understand the above.
I'll do the first one for you- graph it. The conditions are the "if" parts in the piece-wise function. Domain is (-3, inf) and there are no intersepts. Try graphing it. You have a line with slope = 1 and an exponential function.

EDIT: Sorry, there are x and y-intercepts, as Soroban pointed out, although the two graphs do not not intersect which is what I was getting at.

3. Hello, symmetry!

For both questions below:
. . (a) Find the domain of the function.
. . (b) Locate any intercepts.

Did you make a sketch?

$(1)\;\;f(x) \:=\:\begin{Bmatrix} 3 + x & &\text{if }\text{-}3 \leq x < 0 \\ 3 & &\text{if }x = 0 \\ \sqrt{x} & &\text{if }x > 0 \end{Bmatrix}$
Code:
                |
|            *
*      *
* |  *
*   |*
----*-----o--------------
-3     |
Domain: . $(\text{-}3,\,\infty)$

Intercepts: . $(\text{-}3,0),\:(0,3)$

$(2)\;\;f(x)\:=\:\begin{Bmatrix}\frac{1}{x} & & \text{if }x < 0 \\ \sqrt{x} & & \text{if }x \geq 0 \end{Bmatrix}$
Code:
                      |
|            *
|      *
|   *
|*
------------------*----------------
*               |
*         |
*    |
*  |
* |
|
*|
|
Domain: . $(-\infty,\,\infty)$

Intercepts: . $(0,\,0)$

4. ## ok

Thank you again both for your quick replies.

To soroban,

No, I did not sketch the graph because I do not know how to graph piecewise-defined functions.

I understand these functions are graphed in parts, right?

Can you take me through a sample graphing question in terms of this type of function?

Thanks!

5. Originally Posted by symmetry
Thank you again both for your quick replies.

To soroban,

No, I did not sketch the graph because I do not know how to graph piecewise-defined functions.

I understand these functions are graphed in parts, right?

Can you take me through a sample graphing question in terms of this type of function?

Thanks!

Yes, they are 'graphed in parts,' I guess you could call it.

For instance,

Take the first condition;

f(x) = 3 + x if -3 <= x < 0

From x = -3 (including this point) to x = 0 (not including, and thus draw an open circle by this point), you will graph 3 + x; see Soroban's graph. The reason why it's closed (solid dot) is because of the next condition later, and thus includes that point. Try look up piece-wise functions on Wikipedia.

6. ## ok

I like graphing functions. I think piecewise-defined functions are cool but not easy to sketch.

Thanks!

7. Hello again, symmetry!

Okay, here's an example.

. . $f(x) \:=\:\begin{Bmatrix}3 & \text{if }0 \leq x \leq 1 \\ 2x + 1 & \text{if }x > 1\end{Bmatrix}$

When $x$ is between $0$ and $1$ (including the endpoints),
. . the graph is $f(x) = 3$, a horizontal line.
Code:
        |
3* * * * *
|
|
- + - - - + - -
|       1

When $x$ is greater than 1, the graph is $f(x) \:=\:2x + 1$,
. . a slanted line.
Code:
        |
|                   *
|                *
|             *
|          *
3|       *
|
|
- + - - - + - - - - - - -
|       1

Sketch them on the same graph
. . and have the graph of the piecewise function.
Code:
        |
|                   *
|                *
|             *
|          *
3o * * * *
|
|
- + - - - + - - - - - - -
|       1

This function could be your long-distance charge.

They might charge $3 for the first minute . . and$2 per minute for every subsequent minute.

(Hmmm, not a good example . . .
I'm sure someone will point out why.)


# Mellin transform of $x^p$ seems to miss a factor of $2\pi$

Bug introduced in 11.1 or earlier and fixed in 11.3

On Mathematica 11.1.1.0 the Mellin transform of $x^p$ is evaluated as $\delta(p+s)$, while I think it should be $2\pi\,\delta(p+s)$:

In:= MellinTransform[x^p, x, s, GenerateConditions -> True]
Out:= DiracDelta[p + s]

edited posting after Daniel Lichtblau's comment

I initially did not understand this result, but this 2004 paper has explained to me how to arrive at the Dirac delta function, however, with an additional factor of $2\pi$. I checked that this is not a matter of a different definition of the Mellin transform. (I summarized the calculation in this Mathoverflow posting.)

Missing factor $2\pi$ is fixed in Mathematica 11.3.0:

 In:= MellinTransform[x^p, x, s, GenerateConditions -> True]
Out:= 2π DiracDelta[i(p + s)]

consequence: before 11.3 Integrate[MellinTransform[1, x, s], {s, -Infinity, Infinity}] returned 1, now it returns $2\pi\int_{-\infty}^\infty\delta(is)ds$
Q: is this v. 11.3 change in the implementation of MellinTransform documented somewhere?

• See last example in documentation under Scope Elementary Functions. It should be noted that this is a generalization of the integral definition, not unlike the case for FourierTransform. – Daniel Lichtblau Jul 9 '17 at 15:27
• thank you, Daniel, for the feedback, I understand things a bit better now and have edited my posting accordingly --- my problem has been reduced to a missing factor $2\pi$... – Carlo Beenakker Jul 9 '17 at 19:06
• What specific definition is used is not particularly important so long as the MellinTransform and InverseMellinTransform are inverses of each other. Both x^p == InverseMellinTransform[ MellinTransform[x^p, x, s], s, x] and DiracDelta[p + s] == MellinTransform[ InverseMellinTransform[DiracDelta[p + s], s, x], x, s] evaluate to True – Bob Hanlon Jul 9 '17 at 22:41
• @BobHanlon --- but if we assume that the factor of $2\pi$ is absorbed in the definition of DiracDelta, then Integrate[MellinTransform[1, x, s], {s, -Infinity, Infinity}] should return $2\pi$, while instead it returns 1. – Carlo Beenakker Jul 10 '17 at 6:18
• The integral of DiracDelta should be one. – Bob Hanlon Jul 10 '17 at 14:48

The Mellin transforms for $x^j$ reported by Mathematica 11.2 didn't make sense to me, so on 11/28/2017 I submitted the following question on Math StackExchange.

Questions on Mellin Transform of $x^j$ and Interpretation of Distributions with Complex Arguments

I ended up deriving the answer to my own question and on 12/7/2017 I submitted a problem report to Wolfram technical support where I attached a Mathematica notebook illustrating the problem and the correct solution (CASE:3980660).

I received an email from Wolfram technical support on 12/13/2017 indicating my analysis was accepted as correct and a report was being filed with the developers. The correct solution was subsequently implemented in Mathematica 11.3.

Note that not only was the $2\,\pi$ prefix missing, but $i$ was also missing in the $\delta$ function parameter.

I subsequently posted the correct solution in answers to related questions on both Math StackExchange and MathOverflow StackExchange.

Delta function with imaginary argument

Dirac Delta function with a complex argument


# Finding Horizontal Tangent Planes on S

1. Dec 2, 2011

### TranscendArcu

1. The problem statement, all variables and given/known data
S is the surface with equation $$z = x^2 +2xy+2y$$a) Find an equation for the tangent plane to S at the point (1,2,9).
b) At what points on S, in any, does S have a horizontal tangent plane?

3. The attempt at a solution
$$F(x,y,z): z = x^2 +2xy+2y$$
$$F_x = 2x + 2y$$
$$F_y = 2x + 2$$

Evaluated at (1,2) gives answers 6 and 4, respectively. My equation for a plane is:

$$z-9=6(x-1) + 4(y-1)$$.

I think any horizontal plane should have normal vector <0,0,k>, where k is some scalar. I'm pretty sure that S has no such normal vector. But if
$$F(x,y,z): 0 = x^2 +2xy+2y - z$$
then
$$grad F = <2x + 2y,2x + 2,-1>$$ It seems like I can let (x,y) = (-1,1) to zero the x-, y-components of the gradient. Plugging (-1,1) into the definition of z gives z = 1. This suggests to me that there is a point (-1,1,1), at which there is a horizontal tangent plane. Yet I feel pretty sure that this isn't true!

2. Dec 3, 2011

### ehild

You made a little mistake when writing out the equation of the tangent plane. The y coordinate of the fixed point is 2, you wrote 1.

A surface in 3D is of the form F(x,y,z) = constant. For this surface, x2+2xy+2y-z=0. That means F(x,y,z)=x2+2xy+y-z. The gradient of F is normal to the surface, and the tangent plane of the surface at a given point. You want a horizontal tangent plane, so a vertical gradient:(0,0,a). That means Fx=2x+2y=0, Fy=2x+2=0 --->x=-1, y=1, so your result for the x,y coordinates are correct. Plugging into the original equation for x and y, you got z=x2+2xy+2y=1, it is correct. Why do you feel it is not?

ehild

3. Dec 3, 2011

### TranscendArcu

When I graphed F(x,y,z) in MatLab (and it's possible I graphed it incorrectly), I observed that the the resulting paraboloid is always "tilted". Below is a picture from my plot:
http://img440.imageshack.us/img440/687/skjermbilde20111203kl85.png [Broken]
How can this surface have a horizontal tangent anywhere when it is tilted like this?

Last edited by a moderator: May 5, 2017
4. Dec 3, 2011

### ehild

Try to plot z out for -2<x<0 and 0<y<2

ehild

5. Dec 3, 2011

### TranscendArcu

http://img7.imageshack.us/img7/139/skjermbilde20111203kl10.png [Broken]
Hmm. I'm not seeing the a point in this picture where the gradient is pointing directly upwards. Everything still looks kind of tilted.

Last edited by a moderator: May 5, 2017
6. Dec 3, 2011

### ehild

The function is equivalent with z=(x+2y-1)(x+1)+1 and z=1 along the lines x=-1 and y=(1-x)/2. I attach a plot of the surface near the point (-1,1)

ehild

File size:
81.6 KB
Views:
127
7. Dec 4, 2011

### HallsofIvy

Staff Emeritus
I presume that by "horizontal" you mean perpendicular to the z-axis.

The simplest way to find a tangent planes for a surface is to write it in the form F(x,y,z)= constant. Then the normal to the tangent plane at any point is given by $\nabla F$. Here, you can write $F(x, y,z)= x^2+ 2xy+ 2y- z= 0$.

What is $\nabla F$? That will be vertical (and so tangent plane horizontal) when its x and y components are 0.

8. Dec 4, 2011

### ehild

@HallsofIvy: The OP has shown the solution in his first post, he only can not believe it, as the surfaces he got with MatLab look tilted. If you could give advice how to plot surfaces with MatLab, that would be real help for him.

ehild

9. Dec 4, 2011

### TranscendArcu

If anyone is familiar with MatLab, this is the code I've been using:

Note that the "%" mark my annotations. I included them so that hopefully you can follow what I'm doing more easily.

10. Dec 4, 2011

### TranscendArcu

Ha! I figured it out. I forgot a "."

I should have written

z = x.^2 + 2*x.*y +2*y;

Everything makes sense now.

11. Dec 4, 2011

### ehild

You see: it is worth typing something out again and again. Is your plot similar to my one? It was made with Origin. I would like to see your final plot... Please...

ehild

12. Dec 4, 2011

### TranscendArcu

http://img259.imageshack.us/img259/2104/skjermbilde20111204kl10.png [Broken]It looks like it could have a horizontal tangent plane right around (-1,1,1)

Last edited by a moderator: May 5, 2017
13. Dec 4, 2011

### ehild

It is really nice!!! And a missing dot made you sceptical about the truth of Maths??!!!:uhh:

ehild