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Lamcc83
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NP Flash Card Question

by Lamcc83 Tue Feb 11, 2014 1:32 am

I need clarification about question 29 from here:
http://www.manhattangmat.com/pdf/FlashCards_NP_2009.pdf

What is the only two"digit number that is both a perfect square and a perfect cube?

We need a 2-digit integer that is both a perfect square and a perfect cube. This set includes all integers of the form m^3 = n^2, where both m and n are integers. Manipulating the equation tells us that n=m^3/2. Thus we can only choose integers form that will make n an integer"”so m must be a perfect square. The only perfect square that works is 4: 43 = 64, a 2-digit integer. 9 doesn’t work, because 93 = 729, a 3-digit integer. 1 doesn’t work either, because 13 = 1, a 1-digit integer.


How does the manipulation in bold apply to exponents? Which property of exponents states that that manipulation is valid?

Thanks in advance!
RonPurewal
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Re: NP Flash Card Question

by RonPurewal Wed Feb 12, 2014 2:46 am

Lamcc83 Wrote:I need clarification about question 29 from here:
http://www.manhattangmat.com/pdf/FlashCards_NP_2009.pdf

What is the only two"digit number that is both a perfect square and a perfect cube?

We need a 2-digit integer that is both a perfect square and a perfect cube. This set includes all integers of the form m^3 = n^2, where both m and n are integers. Manipulating the equation tells us that n=m^3/2. Thus we can only choose integers form that will make n an integer"”so m must be a perfect square. The only perfect square that works is 4: 43 = 64, a 2-digit integer. 9 doesn’t work, because 93 = 729, a 3-digit integer. 1 doesn’t work either, because 13 = 1, a 1-digit integer.


How does the manipulation in bold apply to exponents? Which property of exponents states that that manipulation is valid?

Thanks in advance!


That step can be seen in two equivalent ways:

1/
Taking the square root of both sides,

2/
Raising both sides to the 1/2 power.

These are the same thing, of course, but you may find one easier to think about than the other.