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Question on Perfect Square 8

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Question on Perfect Square 8

by richachampion » Mon Oct 17, 2016 3:16 am
For any even integer P, 300 multiplied by P is square of an integer. What is the least value of P?

A) 3
B) 4
C) 10
D) 12
E) 14

OA: D
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by richachampion » Mon Oct 17, 2016 3:16 am
Not a tough question, but question stem is slightly Tricky.
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by fiza gupta » Mon Oct 17, 2016 5:51 am
300*p(even integer) = perfect square
let y = 100*3*P
100 is a perfect square but not 3
to make y a perfect square P should be 3 or a multiple of 3
p=3 or p =3*z(z a perfect square)(3*4,3*9.....)
but p is even so
p = 3*4(12),3*16(48)...........

only option D satisfies
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by [email protected] » Mon Oct 17, 2016 10:16 am
Hi richachampion,

This question is based on the same concept as the other prime-factor questions that you posted: when you prime-factor a perfect square, each of the prime factors MUST show up an EVEN number of times...

eg. 9 = (3)(3) here, there are TWO 3s.
eg. 100 = (2)(2)(5)(5) here, there are TWO 2s and TWO 5s
eg. 16 = (2)(2)(2)(2) = here, there are FOUR 2s
Etc.

We're told that P is an EVEN INTEGER and that (300)(P) is a PERFECT SQUARE. We have to be careful to incorporate ALL of the information that we're given.. Prime-factoring this number will get us...

(300)(P) =
(10)(30)(P) =
(2)(5)(2)(3)(5)(P) =
(2)(2)(3)(5)(5)(P)

We have TWO 2s and TWO 5s, but only ONE 3. We need there to be an even number of each prime factor. At first glance, you might think that P must equal 3... However, 3 is NOT an EVEN INTEGER. So we need P to contain a '3' AND and at least TWO 2s (so that P will be EVEN).

Final Answer: D

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by Matt@VeritasPrep » Fri Oct 28, 2016 12:42 am
The OA (and the explanations above) are not quite correct.

Let's set p = 2k, where k is some unknown integer.

300 * p = a square

300 * 2k = a square

600k = a square

(100) * 6k = a square

Since 100 is a square, 6k must also be a square. If k = 0, then 6k is a square, and we're set. Any k < 0 won't work, since squares are nonnegative.

Since p = 2k, p = 0 is our smallest solution, and we're done!

(Unfortunately, whoever wrote this question forgot to put the correct answer. :/ But if we want p > 0, you can see that the next 6k that is a square is 6*6, which gives k = 6 and p = 2*k = 2*6 = 12.)
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by Matt@VeritasPrep » Fri Oct 28, 2016 12:46 am
richachampion wrote:Not a tough question, but question stem is slightly Tricky.
Beware! :D

The question was subtle enough to fool whoever wrote it (p = 0 is an annoying detail, but exactly the sort of annoying detail that the GMAT likes to test!) ... on the GMAT, if it seems easy, be very suspicious!
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by Scott@TargetTestPrep » Mon Oct 31, 2016 10:32 am
richachampion wrote:For any even integer P, 300 multiplied by P is square of an integer. What is the least value of P?

A) 3
B) 4
C) 10
D) 12
E) 14

OA: D
If 300P is a perfect square, then its prime factorization must contain only even exponents. Let's begin by prime factoring 300.

300 = 30 x 10 = 5 x 6 x 5 x 2 = 5 x 3 x 2 x 5 x 2 = 5^2 x 3^1 x 2^2

We can see that 300 is not a perfect square because its prime factorization contains an odd exponent (that is, 3^1). This means that P must contribute the missing factor. However, we are given that P must also be an even integer. Thus, in order to keep "an even number" of twos, P must be divisible by 2^2. Thus P must be divisible by 3^1 x 2^2 = 12. Since we want the least value of P, P must be 12.

In fact, if P = 12, then 300P = 5^2 x 3^2 x 2^4, which is a perfect square.

Answer: D

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