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What is the smallest positive integer K such that the...

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by swerve » Thu Dec 07, 2017 10:35 am
What is the smallest integer K such that the product of 1575 x K is a perfect square?

A. 7
B. 9
C. 15
D. 25
E. 63

The OA is A.

Here's a question I came across and am having problems solving. Any expert that can help me with it, would be greatly appreciated! Thanks.
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by [email protected] » Thu Dec 07, 2017 12:44 pm
Hi swerve,

We're asked to find the SMALLEST integer K such that (1575)(K) is a perfect square. This question can be solved with Prime Factorization.

A 'perfect square' means that we're multiplying an integer by itself. For example:
(2)(2) = 4
(10)(10) = 100

In the second example, we can 'break down' the 10s into prime numbers...
(2x5)(2x5) = 100
The same concept applies here: we have to use the smallest K possible to create two equal terms....

(1575)(K) =
(5)(315)(K) =
(5)(5)(63)(K) =
(5)(5)(7)(9)(K) =
(5)(5)(7)(3)(3)(K)

Based on the above prime factorization, each of the two terms will have to include one 3, one 5 and one 7....
(3x5x7)(3x5xK)
Thus, K the smallest possible value of K is 7.

Final Answer: A

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by GMATWisdom » Thu Dec 07, 2017 3:39 pm
swerve wrote:What is the smallest integer K such that the product of 1575 x K is a perfect square?

A. 7
B. 9
C. 15
D. 25
E. 63

The OA is A.

Here's a question I came across and am having problems solving. Any expert that can help me with it, would be greatly appreciated! Thanks.
here we have to find with how many perfect square numbers i.e. 4,9,16,25,... it is divisible.

any number whose sum of digits is divible by 9 must itself be divisible by 9.

if we add the digits of the number we get 1+5+7+5=18 which is a multiple of 9 so it must be divisible by 9.

dividing with 9 we get

(1575 x k)/9 = 175 x k

any figure which ends in 25, or 50, or 75 or 00 must be divisible by 25

further dividing with 25 we get = 7 x k

therefore k must be equal to 7 to make it a perfect square.

hence 7 is the smallest number to make it a perfect square.

so correct answer is A
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by Matt@VeritasPrep » Thu Dec 07, 2017 4:23 pm
Here's a trick: break 1575K into two identical square roots.

For instance, say we have 36. 36 = 2 * 2 * 3 * 3 = (2 * 3) * (2 * 3), or two identical square roots.

Now let's try that with 1575K.

1575K =>

25 * 63 * K =>

5 * 5 * 3 * 3 * 7 * K =>

(5 * 3 * 7) * (5 * 3 * K)

So if K = 7, we'll have two identical roots. Touchdown!
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by Matt@VeritasPrep » Thu Dec 07, 2017 4:25 pm
Another approach (not quite as good, but you never know what's helpful!): pull the integers roots out of 1575K.

√1575K =>

√1575 * √K =>

√(25*9*7) * √K =>

√25 * √9 * √7 *√K =>

5 * 3 * √7 * √K =>

5 * 3 * √7K

So if √7K = an integer, we're set. If K = 7, then √7K = √49 = 7, an integer, so K = 7 is a solution.
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