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In a perfect square, all the prime constituents have to be raised to an even number. For example, 16 is perfect square which is equal to 2 raised to the power 4. 81 is a perfect square which includes 3 raised to the power 4.

In this example, the prime factors of 1575 are 3, 5, 7. 1575 can be written as a product of all its factors as: \(3^2*5^2*7^1\). So you see, only 7 is having an odd number of powers. if the number gets one more 7, it will can be written as \((3*5*7)^2\), which is a perfect square. Answer is A. 7

If all prime factors are raised to the same even power, then the number is definitely a perfect square. In this example, if 7 is multiplied to 1575, all prime factors become raised to the power 2.

As Bunuel did, I will recommend you to check out the GMAT math book of this forum. To supplement this, you may try out MGMAT number properties. It covers all such required fundamentals in a very effective way.
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What is the smallest positive integer k such that the product 1575*k is a perfect square?

1575 * k = 3 * 3 * 5 * 5 * 7 * k

for perfect square , k should be at least 7. Hence A.
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What is the smallest positive integer K such that the product of 1575 x K is a perfect square?

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

Hello,

Here's a question I came across and am having problems solving. Any help would be greatly appreciated!

Thanks for the help.

\(1575\) = \(3^2\) x \(5^2\) x \(7^1\)

So , we are just 7 short of a perfect square number ..

Hence, correct answer must be (A) 7
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