Competition · AMC preparation · step 4 of 4
AMC 10 · 2020B · #22
Grade 8 number-theoryPick an answer.
AMC 10 2020 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
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.
Substitute a single variable
Substitute x = 2⁵⁰: the divisor becomes N = 2x² + 2x + 1 and the dividend 2²⁰² + 202 = 4x⁴ + 202.
One letter replaces the huge 2⁵⁰ — same problem, friendlier symbols.
8.EE.A.1Solve An Easier Related ProblemSpot the Sophie Germain identity
Spot the Sophie Germain identity: 4x⁴ + 1 = (2x² + 1)² - (2x)² = (2x² + 2x + 1)(2x² - 2x + 1).
Complete the square: 4x⁴ + 1 = (2x² + 1)² - (2x)², then difference of squares.
Completing the square first, then splitting a difference of squares, factors the whole expression.
▸ Why?
A difference of two squares is the two numbers added multiplied by the two subtracted.
▸ Why?
Opening that product sends each piece against each piece, which confirms the factoring exactly.
Show it divides exactly
So 4x⁴ + 1 = N · (2x² - 2x + 1): N divides 4x⁴ + 1 exactly, remainder 0.
The Sophie Germain factorization shows N divides 4x⁴ + 1 cleanly.
6.EE.A.3Identify SubproblemsSplit off the extra 201
Split the dividend: 4x⁴ + 202 = (4x⁴ + 1) + 201; the first piece is ≡ 0, so 4x⁴ + 202 ≡ 201 (mod N).
Peel off the multiple of N; only the leftover 201 remains.
6.NS.B.4Identify SubproblemsCheck the remainder is valid
N = 2¹⁰¹ + 2⁵¹ + 1 is astronomically larger than 201, so the remainder is 201 — choice (D).
201 ≪ N, so 201 is the legitimate remainder.
6.NS.C.7Eliminate PossibilitiesThis 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.
- Substitute a single variable
- Spot the Sophie Germain identity
- Show it divides exactly
- Split off the extra 201
- Check the remainder is valid
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