AMC 10 · 2024 · #5
Grade 6 algebraPick an answer.
AMC 10 2024 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Tool #13 (Convert to Algebra) turns the verbal "flip plus to minus" rule into clean arithmetic: each flip of x subtracts 2x from the original total S = 2500, so the new sum is S - 2 · (sum of flipped terms) and we need this ＜ 0, i.e. the flipped terms must sum to more than 1250. Tool #16 (Change Focus) is the strategic insight that lets us minimize the count — instead of asking "how many flips?", ask "what's the maximum reduction per flip?" and you get the greedy choice 99, 97, 95, …. Tool #6 (Guess and Check) finishes by testing the two adjacent candidate values k = 14 and k = 15 in the closed-form sum.
The first 50 odd numbers sum to 50² = 2500 — pair them as 1 + 99 = 100 for 25 hundreds to confirm.
Pairing the first and last terms (1 + 99 = 100) and walking inward is the same equivalent-expressions trick Grade 6 introduces for arithmetic sums.
6.EE.A.3Convert To AlgebraFlipping +x to -x lowers the total by 2x, so if the flipped numbers add to F the new sum is 2500 - 2F.
Letting F stand for "sum of flipped numbers" turns the verbal rule into a clean expression — the Grade 6 "letters stand for numbers" idea.
6.EE.A.2Convert To AlgebraThe "new sum negative" condition, 2500 - 2F ＜ 0, rearranges to F ＞ 1250.
Solving "strictly negative" for the flipped-sum gives the inequality F ＞ 1250 — a Grade 6 "write/solve inequality of the form x ＞ c" step.
6.EE.B.8Convert To AlgebraTo spend flips efficiently, flip the largest odds first; the k biggest sum to k(100 - k).
Picking the heaviest numbers first is the "do the most with the fewest" extreme-principle move; the arithmetic-series formula is a Grade 6 equivalent-expressions calculation.
6.EE.A.3Count The ComplementTest the boundary: 14 · 86 = 1204 falls short, but 15 · 85 = 1275 clears 1250 — so (B).
Two two-digit multiplications bracket the threshold cleanly — Grade 5 multi-digit multiplication is enough.
5.NBT.B.5Guess And CheckThis AMC 10 problem only needs Grade 6 inequalities (turn "new sum negative" into F ＞ 1250) plus Grade 5 multi-digit multiplication — flip the biggest odd numbers first and you cross the line at flip 15!