Competition · AMC preparation · step 4 of 4
AMC 10 · 2024B · #7
Grade 6 number-theoryPick an answer.
AMC 10 2024 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
The three terms all share 7²⁰²⁴, so the algebra move is to factor it out and inspect the bracket 1 + 7 + 7². That is Tool #13 (Convert to Algebra) in its cleanest form — a numerical sum becomes a product whose individual factors can be checked for divisibility one at a time. Tool #7 (Identify Subproblems) then breaks the divisibility question into two pieces: compute 1 + 7 + 49 = 57, then check whether 57 is divisible by 19. If yes, the whole sum is divisible by 19 regardless of what 7²⁰²⁴ equals, and the remainder is 0.
Factor out the smallest power
Factor the smallest power 7²⁰²⁴ out of all three terms, turning the sum into one product.
Pulling out a common factor turns three messy terms into one — Grade 6 "apply the distributive property to generate equivalent expressions."
Pulling out the smallest power turns three terms into one product.
▸ Why?
A factor shared by every term can be lifted out of the whole expression at once.
▸ Why?
An exponent counts how many times a factor is used, so the smallest power is inside all of them.
Evaluate the bracket
Subproblem A: evaluate the bracket by hand — 1 + 7 + 49 gives 57.
Just plug in 7² = 49 and add — Grade 6 evaluating a numerical expression with whole-number exponents.
6.EE.A.1Identify SubproblemsFactor 57
Subproblem B: 57 is a multiple of 19 because 57 = 3 × 19.
Recognizing 57 = 3 × 19 is Grade 4 "find factor pairs / determine multiples" — the only fact this whole problem really depends on.
4.OA.B.4Identify SubproblemsSubstitute back and read off 19
Put 57 = 3 × 19 back, and 19 becomes a factor of the whole sum, so the remainder is 0.
A number with 19 as a factor leaves remainder 0 on division by 19 — the basic meaning of "multiple of" from Grade 6.
6.NS.B.4Convert To AlgebraThis AMC 10 problem only needs Grade 6 factoring — pull the common power out, check whether 19 divides the small leftover, and you don't need to compute any huge power at all!
- Factor out the smallest power
- Evaluate the bracket
- Factor 57
- Substitute back and read off 19
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