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Difficulty: HardExponents, Roots, and Powers of Integers

If mm and nn are positive integers such that 2m−2n=19202^m - 2^n = 1920, what is the value of m+nm + n?

Answer: 18

Answer

18
To solve 2m−2n=19202^m - 2^n = 1920, factor out 2n2^n to express the left side as 2n(2m−n−1)2^n(2^{m-n} - 1). Prime factorizing 1920 gives 27×152^7 \times 15. Since mm and nn are positive integers with m>nm > n, the term (2m−n−1)(2^{m-n} - 1) is an odd integer. Therefore, the power-of-2 term 2n2^n must equal 272^7, which implies n=7n = 7. The odd term (2m−n−1)(2^{m-n} - 1) must equal 1515, leading to 2m−n=16=242^{m-n} = 16 = 2^4, so m−n=4m - n = 4. Solving for mm gives m=11m = 11. Finally, m+n=11+7=18m + n = 11 + 7 = 18.

Step-by-Step Solution

1
Factor the exponential expression
2n(2m−n−1)=19202^n(2^{m-n} - 1) = 1920
Factoring out the smaller power of 2 separates the expression into a power of 2 and an odd integer multiplier.
2
Determine the prime factorization of 1920
1920=27×151920 = 2^7 \times 15
Prime factorization isolates the highest power of 2 (272^7) from the remaining odd factor (1515).
3
Equate corresponding power-of-2 and odd factors
n=7n = 7 and m−n=4m - n = 4
The even component 2n2^n must equal 272^7, giving n=7n = 7. The odd component 2m−n−12^{m-n} - 1 must equal 1515, so 2m−n=16=242^{m-n} = 16 = 2^4, giving m−n=4m - n = 4.
4
Solve for mm and compute m+nm + n
m=11m = 11 and m+n=18m + n = 18
Adding n=7n = 7 to m−n=4m - n = 4 yields m=11m = 11. The requested sum is m+n=11+7=18m + n = 11 + 7 = 18.

Key Concept

Factoring difference of powers using fundamental exponent rules and equating even/odd prime components.
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Topics

Exponents, Roots, and Powers of Integers

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