AMC 10 · 2020 · #22

Grade 8 number-theory

Archetype Arithmetic Expression Decomposition └ Sub-archetype Fraction / Expression Sum Evaluation

polynomial-factoringsimons-favorite-factoring-trickmodular-arithmeticexponentsdifference-of-squares easier-related-problempattern-recognitionpolynomial-substitution ↑ Prerequisites: polynomial-factoringexponents

📏 Long solution 💡 3 insights

Problem

Find the remainder when the huge number 2²⁰² + 202 is divided by 2¹⁰¹ + 2⁵¹ + 1.

Pick an answer.

(A)

100

(B)

101

(C)

200

(D)

201

(E)

202

AMC 10 2020 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Solve an Easier Related Problem

Tool #9 (Easier Problem): replace 2⁵⁰ by a single variable x. Then the divisor becomes 2x² + 2x + 1 and the dividend 4x⁴ + 202 — much friendlier. Tool #5 (Pattern): recognize the Sophie Germain–style identity 4x⁴ + 1 = (2x² + 2x + 1)(2x² - 2x + 1). Tool #7 (Subproblems): split 2²⁰² + 202 = (2²⁰² + 1) + 201, dispose of the first piece via the identity, and read off the remainder. Tool #13 (Algebra): the algebraic identity is the engine. Tool #3 (Eliminate): the split 202 = 1 + 201 already aligns with choice (D); other choices are sanity-killed.

1STEP 1

Substitute x = 2⁵⁰: the divisor becomes N = 2x² + 2x + 1 and the dividend 2²⁰² + 202 = 4x⁴ + 202.

x = 2⁵⁰: N = 2x² + 2x + 1, dividend = 4x⁴ + 202

2STEP 2

Spot the Sophie Germain identity: 4x⁴ + 1 = (2x² + 1)² - (2x)² = (2x² + 2x + 1)(2x² - 2x + 1).

4x⁴ + 1 = (2x² + 2x + 1)(2x² - 2x + 1)

3STEP 3

So 4x⁴ + 1 = N · (2x² - 2x + 1): N divides 4x⁴ + 1 exactly, remainder 0.

4x⁴ + 1 ≡ 0 (mod N)

4STEP 4

Split the dividend: 4x⁴ + 202 = (4x⁴ + 1) + 201; the first piece is ≡ 0, so 4x⁴ + 202 ≡ 201 (mod N).

4x⁴ + 202 ≡ 201 (mod N)

5STEP 5

N = 2¹⁰¹ + 2⁵¹ + 1 is astronomically larger than 201, so the remainder is 201 — choice (D).

r = 201 → (D)

Answer

201

Sanity. The split 202 = 1 + 201 is the giveaway: had the problem asked for 2²⁰² + 100, the answer would be 99; for +202 it is 201. We can also verify the factorization for a small case: try x = 2, so 4x⁴ + 1 = 65 and (2x² + 2x + 1)(2x² - 2x + 1) = 13 · 5 = 65 ✓. The huge instance is the same algebraic identity, just with x = 2⁵⁰.

💡Key takeaway

This AMC 10 problem only needs the Grade 8 difference-of-squares trick — substitute x = 2⁵⁰ to turn the huge exponents into a small polynomial 4x⁴ + 1 = (2x² + 2x + 1)(2x² - 2x + 1), then 2²⁰² + 202 = (2²⁰² + 1) + 201 leaves remainder 201.

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