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

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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
[Reveal] Spoiler: OA

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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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radi wrote:
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.

Answer: C.
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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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Hi radi,

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.

Final Answer:
[Reveal] Spoiler:
C


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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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New post 19 Jun 2016, 18:09
let p=20
r=10
I and II only

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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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New post 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).

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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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New post 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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New post 30 Oct 2017, 12:59
radi wrote:
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.

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