If 600y is a perfect square (the square of an integer)

600y=90,000=300^2
y=150
l and lll only
D

600= 100*6=2^2*5^2*2*3=2^3*3^1*5^2
For 600y to be the smallest square y should be at least 2*3 compensating for the odd powers
The question is must be not can be
Option A

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Not able to understand as to why option III doesn't work. Prime factorization of 600 yields 2^3*3*5^2. Multiplying by 6 yields even powers but so does multiplying prime factors by 5^2*3^3*2 (y). y is divisible by 30 as well.

Notice that the question asks which of the following MUST be an integer, not COULD be an integer.

If 600y is a perfect square (the square of an integer) and y is a positive integer, then which of the following must also be an integer?

I y/6
II y/42
III y/30

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

For \(600y=2^3*3*5^2*y\) to be a perfect square y must be at least \(2*3=6\). Therefore, only y/6 MUST be an integer.

Answer: A.

Thank you for the explanation Bunuel. Ignored the word 'must' and its implication.

If 600y is a perfect square, then its prime factorization must contain only even exponents. Let’s begin by prime factoring 600.

600 = 6 x 100 = 3 x 2 x 10 x 10 = 3 x 2 x 5 x 2 x 5 x 2 = 2^3 x 3^1 x 5^2

We can see that 600 is not a perfect square because its prime factorization contains two odd exponents (that is, 3^1 and 2^3). Since y is a positive integer, the smallest value of y is 2 x 3 = 6, so that 600y would be 2^4 x 3^2 x 5^2, a perfect square.

Thus, of the roman numerals, only y/6 MUST be an integer.

Answer: A

\(600y = 2^3 * 3^1 * 5^2 * y\)

Since, \(600y\) is a perfect square , least value of \(y = 2*3\)

Thus, y must be divisible by \(6\) ; answer will be (A)

Please pay close attention to the word MUST in the question. It signifies that the number y has to be divisible by 6 for case 1 to be true, by 42 for case 2 to be true and by 30 for case 3 to be true. When we check the positive integers, the smallest value of y we find is 6 (by checking for the values of y starting from 1). Naturally dividing 6 by 42 or 30 will not produce an integer. Hence case 2 and 3 can be rejected on the basis of trial method.

Instead of trial method, we can also mathematically find the least value of y to be 6 as below:
as 600y is a perfect square given in the question, we can write 600y as below
600y= \(2^3*3*5^2*y\)
Clearly we can see that the number 600y will be a perfect square if we multiply the number at least with one 2 and one 3 as in that case it would make the power of 2 and 3 in 600y as even (5's power is already even), this would make it a perfect square (Prime factorisation of any perfect square number will have the prime factors in even powers).

Please press Kudos if you like the explanation.

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