2014 AMC 12A Problem 22

Attempt Problem 22 of the 2014 AMC 12A below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2014 AMC 12A solutions, or check the answer key.

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22.

The number 58675^{867} is between 220132^{2013} and 22014.2^{2014}. How many pairs of integers (m,n)(m,n) are there such that 1≤m≤20121\le m\le2012 and 5n<2m<2m+2<5n+1?5^n\lt2^m\lt2^{m+2}\lt5^{n+1}?
278278
279279
280280
281281
282282
Answer: B
Concepts:exponentsystem of equations
Small hint:
Between consecutive powers 5n5^n and 5n+15^{n+1} there are either two or three powers of 22
Big hint:
The inequality asks for three such powers; if d,td,t count the two-power and three-power gaps, then d+t=867d+t=867 and 2d+3t=20132d+3t=2013
Solution:
Because 22<5<23,2^2\lt5\lt2^3, each interval (5n,5n+1)(5^n,5^{n+1}) contains either two or three powers of 2.2. The chain 5n<2m<2m+2<5n+15^n\lt2^m\lt2^{m+2}\lt5^{n+1} holds exactly when the interval contains three consecutive powers of 2,2, and then there is a unique such m.m.
Let dd and tt be the numbers of intervals (5n,5n+1)(5^n,5^{n+1}) for 0≤n≤8660\le n\le866 containing two and three powers of 2,2, respectively. Since 22013<5867<220142^{2013}\lt5^{867}\lt2^{2014} there are 20132013 powers of 22 in total, giving d+t=867d+t=867 and 2d+3t=2013.2d+3t=2013.
Solving, t=2013−2⋅867=279.t=2013-2\cdot867=279.
Thus, the correct answer is B.
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