NandishSS wrote:M is a positive integer less than 100. When m is raised to the third power, it becomes the square of another integer.
How many different values could m be?
A. 7
B. 9
C. 11
D. 13
E. 15
OA: B
What is the concept testing here?
I think the language here could be a little cleaner, but the idea is that if a value is both a perfect square and a perfect cube, then some base has been raised to both the 2 and the 3, meaning that the value in question, when reduced to its prime factorization, will have exponents that are multiples of 6.
So start by listing out the positive perfect squares less than 100: 1, 2^2, 3^2, 4^2, 5^2, 6^2, 7^2, 8^2, 9^2.
Well, if we raised each of these to the third, they'll all still be perfect squares, right?
1^3 = 1 = perfect square.
(2^2)^3 = 2^6 = still a perfect square. (2^6 = (2^3)^2) Put another way, any number raised to an even exponent must be a perfect square.
(3^2)^3 = 3^6 = (3^3)^2 still a perfect a square.
etc.
So all 9 perfect squares less than 100, when cubed, will be the square of some number. The problem with the wording is that it makes it sound as though every number in the list will be the square of a
different number. This is clearly not true of 1. However, the other 8, when cubed, will be the square of another number. (4, or 2^2, when cubed, becomes 64, or 8^2, etc.)
