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

Question

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

JeffTargetTestPrep · Accepted Answer

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

Kurtosis · Answer

({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

hotshot02 · Answer

Vyshak  please can you explain this as I am not able to understand.

Thanks!

Kurtosis · 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.

Mahmud6 · Answer

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.

Luckisnoexcuse · Answer

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

sahilvijay · Answer

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

RMD007 · Answer

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.

mahu101 · Answer

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.”)

RMD007 · Answer

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

mahu101 · Answer

Yes, you are right.

EgmatQuantExpert · Answer

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

bumpbot · Answer

Automated notice from GMAT Club BumpBot:

A member just gave Kudos to this thread, showing it’s still useful. I’ve bumped it to the top so more people can benefit. Feel free to add your own questions or solutions.

This post was generated automatically.

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

Prep Toolkit

Top 5 GMAT Debrief Videos

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards

GMAT Problem Solving (PS) Questions

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

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards