m + n

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m + n

by DanaJ » Tue Dec 07, 2010 10:41 am
Source: Veritas Prep

If m and n are integers and 36/(3^4) =1/(3^m) +1/(3^n) What is the value of m + n?

A. -2
B. 0
C. 2
D. 3
E. 5

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by Night reader » Tue Dec 07, 2010 11:34 am
DanaJ wrote:Source: Veritas Prep

If m and n are integers and 36/(3^4) =1/(3^m) +1/(3^n) What is the value of m + n?

A. -2
B. 0
C. 2
D. 3
E. 5

Experts: only Veritas Prep experts, please!
IOM D 3

4/(3^2)=(3^-m) + (3^-n) => (3^0+3^1)/(3^2)=(3^-m) + (3^-n) => (3^-2)+ (3^-1)= (3^-m) + (3^-n) => m+n=3

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by David@VeritasPrep » Wed Dec 08, 2010 5:29 pm
Correct answer: (D)

Solution: While this problem appears to be a typical exponent problem that you manipulate algebraically, it is best solved with basic arithmetic concepts. On the left side, 36/3^4 can be reduced to 4/9 so that the original equation now reads 4/9=1/3^m +1/3^n Using logic and an understanding of fractions, you can now deduce the values of m and n. Neither n nor m could be 0 or negative as that would make one of the terms on the right greater than or equal to 1 and the sum could never equal the fraction 4/9. Since the lowest common denominator has to be nine, consider the different positive values that add to 4/9. The only possibilities are 1/9+ 3/9 or 2/9+ 2/9. Only the first can be written with 1s in the numerator and 3s in the denominator. 1/9+3/(9 ) can be expressed in 3rds as 1/9+1/3, or 1/3^2 +1/3^1 . Therefore it must be true that either m = 2 and n = 1, or n = 1 and m = 2. Since they are asking for the sum, it does not matter which is which. The answer is 2 +1, or 3.
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