2020 AMC 12A Problem 19

Attempt Problem 19 of the 2020 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 2020 AMC 12A solutions, or check the answer key.

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19.

There exists a unique strictly increasing sequence of nonnegative integers a1<a2<⋯<aka_1 \lt a_2 \lt \cdots \lt a_k such that 2289+1217+1=2a1+2a2+⋯+2ak.\frac{2^{289} + 1}{2^{17} + 1} = 2^{a_1} + 2^{a_2} + \cdots + 2^{a_k}. What is k?k?
117117
136136
137137
273273
306306
Answer: C
Concepts:number basepower of 2factoring
Small hint:
Write x=217;x = 2^{17}; then the quotient is x17+1x+1=x16−x15+⋯+1\dfrac{x^{17} + 1}{x + 1} = x^{16} - x^{15} + \cdots + 1
Big hint:
Group the alternating terms into pairs xm+1−xm=217m(217−1),x^{m+1} - x^m = 2^{17m}(2^{17} - 1), each a block of 1717 ones in binary
Solution:
Let x=217.x = 2^{17}. Then 2289+1217+1\dfrac{2^{289} + 1}{2^{17} + 1} =x17+1x+1= \dfrac{x^{17} + 1}{x + 1} =x16−x15+⋯−x+1,= x^{16} - x^{15} + \cdots - x + 1, an alternating sum of the 1717 powers x0,x1,…,x16.x^0, x^1, \ldots, x^{16}.
Pair each subtracted power with the added power just above it: xm+1−xmx^{m+1} - x^m =217m(217−1)= 2^{17m}(2^{17} - 1) =217m+217m+1= 2^{17m} + 2^{17m+1} +⋯+217m+16,+ \cdots + 2^{17m+16}, a block of 1717 consecutive powers of 2.2.
There are 88 such pairs, together with the leftover +20.+2^0. The blocks occupy disjoint ranges, so the total number of powers is 8⋅17+1=137.8 \cdot 17 + 1 = 137.
Thus, C is the correct answer.
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