Competition · AMC preparation · step 4 of 4
AMC 10 · 2024A · #4
Grade 6 arithmeticnumber-theoryPick an answer.
AMC 10 2024 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
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.
Set up the inequality
Subproblem A: each term is at most 99, so n terms reach at most 99n — to hit 2024 we need 99n ≥ 2024.
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.
The requirement becomes a plain statement that the count must be at least a certain value.
▸ Why?
Each term is capped, so a total needs enough terms to reach the target and no fewer.
▸ Why?
How many full terms fit is exactly a division, and any leftover forces one more term.
Divide 2024 by 99
Divide: 99 × 20 = 1980 is short of 2024, but 99 × 21 = 2079 clears it — so n ≥ 21.
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 CheckCheck that 21 works
Subproblem B: is n = 21 reachable? Write 2024 = 20 × 99 + r, giving r = 2024 − 1980 = 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 SubproblemsBuild the example
Twenty 99s plus one 44 gives exactly 21 two-digit terms summing to 2024; the bound forbids fewer, so the minimum is 21 (B).
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 10's opening problems often reward this Grade 6 inequality + check move.
- Set up the inequality
- Divide 2024 by 99
- Check that 21 works
- Build the example
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