AMC 10 · 2024 · #3
Grade 6 arithmeticnumber-theoryPick an answer.
An "at least" minimum problem really hides two questions, so Tool #7 (Identify Subproblems) splits the work cleanly. Subproblem A: how few terms could possibly work? Each term is at most 99, so n terms can reach at most 99n — that gives a lower bound on n. Subproblem B: can that lower bound actually be achieved with valid two-digit numbers? We pair this with Tool #6 (Guess and Check) to test the boundary cases n = 20 and n = 21 — one fails, one works, and that pins down the answer.
Bound the count
With n terms the sum is at most 99n.
Writing the constraint as an inequality is exactly the Grade 6 move "n ≥ c": n has to be at least as big as 2024/99.
Writing the requirement as an inequality says the count has to be at least the target divided by the largest term.
▸ Why?
Splitting the target by the largest term leaves a quotient and a remainder, and the quotient is the floor on the count.
▸ Why?
Fewer terms could never reach the target, since even the largest terms would fall short.
Check the boundary
Twenty falls short, twenty-one suffices.
Grade 6 fluent multi-digit arithmetic: just multiply 99 by 20 and 21 to see which one first crosses 2024.
6.NS.B.2Guess And CheckCompute the remainder
Twenty 99s leave 44.
After using twenty 99s, we still need 44 more — and 44 is itself a two-digit number, which is the perfect 21st term.
6.NS.B.2Identify SubproblemsExhibit the example
Twenty 99s and one 44 make 21 terms.
The candidate value n = 21 passes the check (a valid sum exists), so the inequality bound is tight — that's how Grade 6 "is this n a solution?" reasoning closes the problem.
6.EE.B.5Guess And CheckAn "as few as possible" problem becomes easy when you ask "as big as possible per piece?" first — the AMC 12's opening problems often reward this Grade 6 inequality + check move.