AMC 10 · 2023 · #5
Grade 6 arithmeticPick an answer.
AMC 10 2023 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
The problem packs two facts into one paragraph; Tool #7 (Subproblems) splits them: (i) Lara's clue alone tells us the original sum S, and (ii) Maddy's clue then tells us how many numbers added up to give that increase. Tool #11 (Work Backwards) applies in step (i) — Lara's sum was tripled to reach 45, so undo the triple to recover the original. A Tool #9 (Easier Problem) sanity check — "what if the list were {5,5,5}?" — reassures that the structure is right. Full algebra (#13) would work but is overkill for two one-step inversions.
Lara's clue: tripling every number triples the total, so 3S = 45; divide by 3 to get S = 15.
Lara's 3-times rule is just one undo away from the original — Grade 3 multiplication/division reverses cleanly.
3.OA.B.5Work BackwardsMaddy's clue: adding 3 to each of the n numbers adds 3n total, so her sum is S + 3n = 45.
"3 added to each of n numbers" totals 3n extra — a Grade 6 expression that records the bookkeeping.
6.EE.A.2Identify SubproblemsSubstitute S = 15 into Maddy's equation and subtract 15 from both sides: 3n = 30.
Once we know the original sum, the extra 30 in Maddy's pile is pure 3-per-number bonus.
6.EE.B.7Work BackwardsDivide by 3 to get n = 10 — ten numbers on the blackboard, choice (A).
If 30 extra came from +3 per number, then 30 ÷ 3 = 10 numbers — Grade 3 division-as-unknown-factor.
3.OA.B.6Identify SubproblemsThis AMC 10 problem only needs Grade 6 expression-and-equation thinking — Lara's clue pins the original sum at 15, and Maddy's extra 30 is just +3 per number, so there are 10 numbers.