M is a positive integer less than 100. When m is raised to the third

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 the perfect square 8^2. Another illustration: Let’s let m = 25 = 5^2. Now, let’s raise m to the third power, obtaining m^3 = (5^2)^3 = 25^3 = 15625, which is the perfect square of 125.

(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
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\(({a^2})^3 = a^6 = ({a^3})^2\)

Given: M = \(0 < a^2 < 100\)

Values of a^2 can be --> \(1^2, 2^2, 2^4, 2^6, 3^2, 3^4, 5^2, 7^2, (2^2 * 3^2)\)

There are 9 possible values.

Answer: B

Ans is B

M^3 = y^2
M= y^(2/3)
means y should have a cube-root as integer and whose square should be less than 100
lets say start by 1000 =y
y^(2/3) = 1000^2/3 =100 rejected means M cant be 100
M can attain 81,64,49,36,25,16,09,04,01 TOTAL 9 values .

Therefore B is the answer

You have to find the number of values of a^2 that are between 0 and 100. --> a is between 0 and 10 --> We will have 9 values for m.

Hope it helps.

Yes, you are right.

Given

• M is a positive integer less than 100.
• When m is raised to the third power, it becomes the square of another integer.

To Find

• The number of possible values of m.

Approach and Working Out

• When m is raised to third power it becomes m^3.

o This is also a square so we can say m must be a perfect square.
o Please note, here m does not become a perfect square due to raising it to power 3, it is already a perfect square.

• We need to find the number of perfect squares less than 100.

o That is 9.

Correct Answer: Option B
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Vyshak please can you explain this as I am not able to understand.

Thanks!

Why we need to look for perfect squares that are less than 100? We need the values of \(m<100\) NOT the \(m^2<100\).

I know I am wrong, but I don't know why I am wrong.
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0<x^2<100
1,4,9,16,25,36,49,64,81
9 numbers
B


Sent from my iPhone using GMAT Club Forum mobile app
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Bunuel,
For this question, I understood that basically we need to find out squares of integers below 100. But I have one query -

Why does above answers includes \(1^2\)? The question stem says if the number is raised to third power, then it becomes square of ANOTHER integer?

\(\sqrt{(1^3)}\) = 1 only.

Please explain.

JeffTargetTestPrep addressed this in his response above -

(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.”)

Thanks, my point is, with the given question stem 8 should be the answer!

Hello JeffTargetTestPrep
When you say "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. "
is there any rule or axiom that any perfect square raised to say 3rd or any power will always give a perfect square?

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