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2020 AMC 12A Problem 19

Problem 19 of 25HarderAlgebraNumber Theory

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?

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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} =x16x15+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+1xmx^{m+1} - x^m =217m(2171)= 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 817+1=137.8 \cdot 17 + 1 = 137. Thus, C is the correct answer.

More practice

Concepts: number base · power of 2 · factoring

Problem text and solution from the LIVE past-contest archive. See also the AoPS wiki page for community solutions.