AMC 10 · 2024 · #8
Grade 8 arithmeticPick an answer.
AMC 10 2024 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
The pattern Tool #5 spots is the symmetry of divisors: every divisor d has a unique "partner" , and the two multiply to 42. So once we list the divisors systematically (Tool #2), we can pair them into 42 × 42 × 42 × 42 = 42⁴. Tool #7 (Identify Subproblems) then reduces the units-digit question to just "what is the ones digit of 42⁴?" — which depends only on the ones digit 2 raised to the 4th power.
Since 42 = 2 × 3 × 7, every divisor is a subset-product of {2, 3, 7}, giving 8 divisors in all.
Listing every divisor in order is exactly the Grade 4 "find all factor pairs" move — every divisor has to show up here.
4.OA.B.4Make A Systematic ListMatch each small divisor with the partner that multiplies to 42; the eight divisors split into 4 such pairs.
Divisors come in pairs whose product is the original number — Tool #5 spots this symmetry and Grade 4 "factor pair" thinking confirms it.
4.OA.B.4Look For A PatternEach of the 4 pairs contributes a factor of 42, so the whole product collapses to 42⁴.
Four pairs each equaling 42 multiply to 42⁴ — Grade 6 "evaluate expressions with whole-number exponents".
6.EE.A.1Identify SubproblemsA power's ones digit tracks the base's ones digit, so 42⁴ ends in the ones digit of 2⁴ = 16, namely 6 — choice (D).
Tens-and-higher digits cannot affect the ones digit of a product — Grade 8 integer-exponent properties make this routine.
8.EE.A.1Identify SubproblemsThis AMC 10 problem only needs Grade 8 exponent rules — pair the divisors of 42 so each pair multiplies to 42, and the ones digit of 42⁴ is the ones digit of 2⁴ = 16!