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.)

(Apologies to Matt for the excessive '^' usage. I can't help myself.)

HI Dave,

One quick question, Suppose 11^2 when raised to 11^3 then 11^3 is square of some number. Am I right?

Thanks
Nandish

NandishSS wrote:HI Dave,

One quick question, Suppose 11^2 when raised to 11^3 then 11^3 is square of some number. Am I right?

Thanks
Nandish

If you mean the case when 11^2 is cubed to give you (11^2)^3 = 11^6, then the answer is yes, 11^6 is 11^3 squared, or (11^3)^2.

Just remember that this question specifies numbers that are less than 100 before they're cubed, so 11'2 wouldn't be eligible here.

Here's another question that tests the concept in a similar way:

https://www.beatthegmat.com/n-a-3b-4c-5- ... tml#767646

m³ = n²

So we're looking for numbers that are both cubes and squares.

If a number is a cube, then it has some multiple of 3 of each of its prime factors.

If a number is a square, then it has some multiple of 2 of each of its prime factors.

If a number is both a cube AND a square, then it has some multiple of (2 * 3) of each of its prime factors. Such a number must be a sixth power.

When we cube our number m, we'll automatically get 3 of each of m's prime factors, so the cube part is set. What we then need is for each of m's primes to become even when multiplied by 3: in other words, for m to ALREADY have an even number of each of its primes.

That's a fancy way of saying that we need m to be a square. The positive squares less than 100 are 1, 4, 9, 16, 25, 36, 49, 64, and 81, so we've got 9 of them. (I actually think the wording ("another integer") isn't perfectly clear and might disqualify 1 as a solution, but since eight isn't an answer choice, we have to go with nine.)

DavidG@VeritasPrep wrote:(Apologies to Matt for the excessive '^' usage. I can't help myself.)

Fear not, help is on the way! Superscripts Anonymous is always in session here. (If you forget the codes, just bookmark the link and copypaste the relevant superscript for each problem.)

The only ^s we should see on this board are in goofy ASCII smiley faces perpetrated by yours truly! (〃^∇^)ﾉ

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

We are given that m is a positive integer less than 100. We are also given that when m is raised to the third power, it becomes the square of another integer. In order for that to be true, m itself must (already) be a perfect square, since any perfect square raised to the third power will still be a perfect square, i.e., square of an integer. Thus we are looking for perfect squares that are less than 100. Since there are 9 perfect squares that are less than 100, namely, 1, 4, 9, ..., 64, and 81, the answer is 9.

Let's look at some examples to clarify this: Let's assume that m = 4 = 2^2. Now, let's raise m to the third power, obtaining m^3 = (2^2)^3 = 4^3 = 64, which is 8^2. Another illustration: let m = 25 = 5^2. Now, let's raise m to the third power, obtaining m^3 = (5^2)^3 = 25^3 = 15,625, which is 125^2.

(Note: By the way the problem is worded, "when m is raised to the third power, it becomes the square of another integer," 1 should not be counted as one of the 9 different values m could be, unlike all the other 8 values. For example, take the number 4: 4^3 = 64 = 8^2, which is the square of another integer, 8. However, 1^3 = 1 = 1^2, which is the square of the same integer. The correct way to word the problem is "when m is raised to the third power, it becomes the square of an integer.")

Answer: B