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by Joy Shaha » Fri May 06, 2016 11:43 pm
Q. What is the greatest positive integer n such that (2^n) is a factor of (12^10)?
A. 10 ; B. 12 ; C. 16 ; D. 20 ; E. 60
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by [email protected] » Sat May 07, 2016 9:14 am
Hi Joy Shaha,

This question comes down to exponent rules and how to 'rewrite' a number raised to a power.

12 = (2)(2)(3)

12^2 = (2^2)(2^2)(3^2) = (2^4)(3^2)
12^3 = (2^3)(2^3)(3^3) = (2^6)(3^3)

Seeing this pattern, you should be able to rewrite 12^10...

12^10 = (2^10)(2^10)(3^10) = (2^20)(3^10)

So the largest 'power of 2' that is a factor of 12^10 is 20.

Final Answer: D

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by 800_or_bust » Sat May 07, 2016 11:16 am
Joy Shaha wrote:Q. What is the greatest positive integer n such that (2^n) is a factor of (12^10)?
A. 10 ; B. 12 ; C. 16 ; D. 20 ; E. 60
So 12^10 is simply mathematical shorthand to express that we have ten 12's. Each 12 can be factored into two 2's and one 3. Therefore, we ten separate pairs of two 2's and one 3. This means 12^10 can be expressed as ((2^2)x3)^10, which is the same as 2^20 x 3^10. Therefore, the answer is D.
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by OptimusPrep » Sat May 07, 2016 7:31 pm
Joy Shaha wrote:Q. What is the greatest positive integer n such that (2^n) is a factor of (12^10)?
A. 10 ; B. 12 ; C. 16 ; D. 20 ; E. 60
The question asks for what greatest value of n is 2^n a factor on 12^10

12^10 = 4^10*3^10 = 2^20*3^10
This means there are 20 powers of 2's in 12^10

Hence maximum value of 2^n can be 2^20 for it to be a factor.
Hence n = 20

Correct Option: D
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by Matt@VeritasPrep » Thu May 12, 2016 12:25 am
12¹� = (2² * 3)¹� = 2²� * 3¹�.

The exponent on 2 is 20, so D.
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