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# If p and r are positive integers such that (p^2)/40=r, and p ≠ r, then

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Joined: 22 Jul 2011
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If p and r are positive integers such that (p^2)/40=r, and p ≠ r, then  [#permalink]

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25 Feb 2015, 10:56
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If p and r are positive integers such that $$\frac{(p^2)}{40}=r$$, and p ≠ r, then which of the following must also be an integer?

I. $$\frac{r}{5}$$
II. $$\frac{r}{(2*5)}$$
III. $$\frac{r}{(3*5)}$$

(A) I only
(B) II only
(C) I and II only
(D) I and III only
(E) I, II, and III
Math Expert
Joined: 02 Sep 2009
Posts: 51072
Re: If p and r are positive integers such that (p^2)/40=r, and p ≠ r, then  [#permalink]

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25 Feb 2015, 11:15
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If p and r are positive integers such that $$(p^2)/40=r$$, and p=/=r, then which of the following must also be an integer?

I. $$r/5$$
II. $$r/(2x5)$$
III. $$r/(3x5)$$

(A) I only
(B) II only
(C) I and II only
(D) I and III only
(E) I, II, and III

(p^2)/40 = r;

p^2 = 40r = 2^3*5r.

2^3*5r equals to the square of an integer (p^2), thus r should complete the powers 2^3 and 5 to the even number, thus the least value of r is 2*5 (in this case p^2 = 2^3*5r = 2^4*5^2). So, r/5 and r/(2*5) must be also be integer.

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Re: If p and r are positive integers such that (p^2)/40=r, and p ≠ r, then  [#permalink]

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25 Feb 2015, 15:59
3
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The information in this prompt can help you to make some deductions about P and R. We're told that they're both positive integers and that (P^2)/40 = R. The restriction that P cannot equal R means that P and R CANNOT equal 40. That restriction actually has no bearing on this prompt.

To start, we can get rid of the fraction:

P^2 = 40(R)

P^2 = (P)(P), so each of those parentheses MUST contain the SAME prime factors.

40(R) = (2)(2)(2)(5)(R)

Since there are two Ps, for each "P" to contain the same primes, we'll need an EVEN number of 2s and an EVEN number of 5s.

So far, we have three 2s, so we'll need AT LEAST one more 2. We also have just one 5, so we'll need AT LEAST another 5. Both that "extra" 2 and "extra" 5 MUST be in the R....

IF...R = 10, then 40(R) = (40)(10) = (2)(2)(2)(2)(5)(5) and P^2 = (2)(2)(5) = 20

This all proves 2 things:
1) R MUST be a multiple of 10 (but cannot be 40)
2) At the minimum, P = 20

Using this information, we can now deal with the Roman Numerals. Which of the following MUST be INTEGERS....?

I. R/5

Since R must be a multiple of 10, we know that R/5 is an integer.
Roman Numeral I must be an integer.

II. R/[(2)(5)]

Again, since we know that R must be a multiple of 10, R/10 is an integer.
Roman Numeral I must be an integer.

III. R/[(3)(5)]

R/15 may or may not be integer, depending on the value of R.
IF....R = 10, then R/15 is NOT an integer
IF....R = 30, then R/15 IS an integer.
Roman Numeral III is NOT necessarily an integer.

GMAT assassins aren't born, they're made,
Rich
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# Rich Cohen

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Special Offer: Save $75 + GMAT Club Tests Free Official GMAT Exam Packs + 70 Pt. Improvement Guarantee www.empowergmat.com/ *****Select EMPOWERgmat Courses now include ALL 6 Official GMAC CATs!***** VP Joined: 07 Dec 2014 Posts: 1128 Re: If p and r are positive integers such that (p^2)/40=r, and p ≠ r, then [#permalink] ### Show Tags 19 Jun 2016, 18:09 let p=20 r=10 I and II only EMPOWERgmat Instructor Status: GMAT Assassin/Co-Founder Affiliations: EMPOWERgmat Joined: 19 Dec 2014 Posts: 13057 Location: United States (CA) GMAT 1: 800 Q51 V49 GRE 1: Q170 V170 Re: If p and r are positive integers such that (p^2)/40=r, and p ≠ r, then [#permalink] ### Show Tags 20 Jun 2016, 09:14 Hi gracie, While you've found one set of values that confirms the correct answer, you should be a bit more careful when dealing with similar questions in the future. When a Roman Numeral question asks for what MUST be an integer (or MUST be true, etc.), that really means "what MUST be an integer every single time no matter how many different examples you come up with..." TESTing VALUES can be quite useful on these types of questions, but you have to acknowledge that the first TEST that you come up with is not the only possible TEST (and an alternative TEST might prove that the calculation is not an integer every time). On a concept level, it's similar to how DS questions are designed - you're looking to prove whether the answer to the question is consistent or inconsistent (and one example doesn't prove it either way). GMAT assassins aren't born, they're made, Rich _________________ 760+: Learn What GMAT Assassins Do to Score at the Highest Levels Contact Rich at: Rich.C@empowergmat.com # Rich Cohen Co-Founder & GMAT Assassin Special Offer: Save$75 + GMAT Club Tests Free
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Intern
Joined: 04 Oct 2017
Posts: 12
Re: If p and r are positive integers such that (p^2)/40=r, and p ≠ r, then  [#permalink]

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13 Oct 2017, 05:55
Can it not alao be like the following?
P^2=40R
P= 40R^1/2
P= 2(10R)^1/2
For P to be integer 10R has to be an integer. This is satisfied by R=10
So from the options only I & II gives resultant as an integer.

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Re: If p and r are positive integers such that (p^2)/40=r, and p ≠ r, then  [#permalink]

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30 Oct 2017, 12:59
If p and r are positive integers such that $$\frac{(p^2)}{40}=r$$, and p ≠ r, then which of the following must also be an integer?

I. $$\frac{r}{5}$$
II. $$\frac{r}{(2*5)}$$
III. $$\frac{r}{(3*5)}$$

(A) I only
(B) II only
(C) I and II only
(D) I and III only
(E) I, II, and III

Since 40 = 2^3 x 5^1 and r is a positive integer, the smallest positive integer p such that p^2 is divisible by 40 is 2^2 x 5 = 20. Thus, p^2 = 2^4 x 5^2 = 400 and 400/40 = 10 = r.

Since 10 is divisible by 5 and (2 x 5) but not by (3 x 5), I and II are true.

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Re: If p and r are positive integers such that (p^2)/40=r, and p ≠ r, then &nbs [#permalink] 30 Oct 2017, 12:59
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