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

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M is a positive integer less than 100. When m is raised to the third [#permalink]

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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


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Re: M is a positive integer less than 100. When m is raised to the third [#permalink]

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New post 06 Oct 2016, 20:13
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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

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Re: M is a positive integer less than 100. When m is raised to the third [#permalink]

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New post 01 Mar 2017, 12:35
Vyshak wrote:
\(({a^2})^3 = a^6 = ({a^3})^2\)

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


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

Thanks!
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Re: M is a positive integer less than 100. When m is raised to the third [#permalink]

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hotshot02 wrote:

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

Thanks!


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.

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Re: M is a positive integer less than 100. When m is raised to the third [#permalink]

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idontknowwhy94 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


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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Re: M is a positive integer less than 100. When m is raised to the third [#permalink]

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New post 26 Aug 2017, 01:14
JeffTargetTestPrep wrote:
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.

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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Re: M is a positive integer less than 100. When m is raised to the third [#permalink]

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New post 26 Aug 2017, 02:09
idontknowwhy94 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


Keep the Kudos dropping in and let these tricky questions come out .... :lol: :lol:

0<x^2<100
1,4,9,16,25,36,49,64,81
9 numbers
B


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Re: M is a positive integer less than 100. When m is raised to the third [#permalink]

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New post 10 Sep 2017, 07:38
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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
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Re: M is a positive integer less than 100. When m is raised to the third [#permalink]

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New post 21 Oct 2017, 22:01
idontknowwhy94 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


Keep the Kudos dropping in and let these tricky questions come out .... :lol: :lol:


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.
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Re: M is a positive integer less than 100. When m is raised to the third [#permalink]

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New post 21 Oct 2017, 23:09
RMD007 wrote:
idontknowwhy94 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


Keep the Kudos dropping in and let these tricky questions come out .... :lol: :lol:


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.”)

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Re: M is a positive integer less than 100. When m is raised to the third [#permalink]

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New post 21 Oct 2017, 23:13
mahu101 wrote:

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!
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Re: M is a positive integer less than 100. When m is raised to the third [#permalink]

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New post 21 Oct 2017, 23:19
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RMD007 wrote:
mahu101 wrote:

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!


Yes, you are right. :-)

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Re: M is a positive integer less than 100. When m is raised to the third   [#permalink] 21 Oct 2017, 23:19
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