AMC 10 · 2020 · #19

Grade 8 number-theory
polynomial-factoringbase-conversionexponentssequences-geometricpattern-recognition identify-subproblemspattern-recognitioneasier-related-problem ↑ Prerequisites: polynomial-factoringbase-conversion
📏 Long solution 💡 3 insights
Problem
Write a certain large integer as a sum of distinct powers of two with strictly increasing exponents. Find how many terms there are — that is, how many ones appear in its binary representation.

Pick an answer.

(A)
117
(B)
136
(C)
137
(D)
273
(E)
306
How to solve
Strategy Solve an Easier Related Problem

Tool #9 (Easier Problem): the exponents 289 and 17 are huge — replace them with the substitution x = 2¹⁷ to reveal a cleaner (x¹⁷+1)/(x+1), then drop further to the tiny prototype (x³+1)/(x+1) with x=2² to see the mechanism. Tool #5 (Pattern): the alternating polynomial quotient x¹⁶ - x¹⁵ + … + 1 is a clean pattern. Tool #7 (Subproblems): split each subtraction 2^A - 2^B into a string of 1-bits. Tool #2 (Systematic List): once we know each pair contributes 17 bits, list the pair-blocks and sum.

1STEP 1

Substitute to simplify

Substituting gives a familiar polynomial division.

(2²⁸⁹+1)/(2¹⁷+1) = (x¹⁷+1)/(x+1), x = 2¹⁷
2STEP 2

Write the quotient

The quotient alternates in sign.

(x¹⁷+1)/(x+1) = x¹⁶ - x¹⁵ + x¹⁴ - … + x² - x + 1
3STEP 3

Return to powers of two

Undo the substitution.

2²⁷² - 2²⁵⁵ + 2²³⁸ - 2²²¹ + … + 2³⁴ - 2¹⁷ + 2⁰
4STEP 4

Pair the neighbours

Pair neighbours to kill the negatives.

(2²⁷²-2²⁵⁵)_pair 1 + (2²³⁸-2²²¹)_pair 2 + … + (2³⁴-2¹⁷)_pair 8 + 1
5STEP 5

Expand one pair

Each pair becomes a run of consecutive bits.

2^A - 2^B = 2^B(2¹⁷-1) = 2^A-1 + 2^A-2 + … + 2^B+1 + 2^B
6STEP 6

Count the blocks

Count the blocks and their width.

blocks: [17,33], [51,67], [85,101], …, [255,271]; 8 blocks of width 17
7STEP 7

Add it all up

Including the final one gives 137.

k = 8 × 17 + 1 = 136 + 1 = 137 → (C)
Answer
137
Magnitude check: the quotient is about 2²⁸⁹/2¹⁷ = 2²⁷², which has ≤ 273 bits total. Among the 273 possible bit positions, k = 137 are 1 — roughly half, which matches the alternating-then-fill pattern. Choice (D) 273 is the bit-length, a trap; (B) 136 misses the lone 2⁰; (C) 137 is correct. Spot-check with the tiny case (2⁹+1)/(2³+1) = 513/9 = 57 = 32 + 16 + 8 + 1 = 2⁵ + 2⁴ + 2³ + 2⁰, which has k=4 bits. Using the same formula: 9 = 3 · 3, alternating polynomial has 3 terms (1 pair + trailing 1), so k = 1 · 3 + 1 = 4. Matches.
💡Key takeaway

This AMC 12 problem only needs Grade 8 exponent rules you already know: rename x = 2¹⁷ to shrink the fraction to (x¹⁷+1)/(x+1), expand into 17 alternating powers of 2, pair them up so each pair becomes 17 consecutive 1-bits in binary, and add: 8 × 17 + 1 = 137.