AMC 10 · 2019 · #5
Grade 6 arithmeticPick an answer.
AMC 10 2019 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Tool #9 (Easier Problem): try a smaller target sum first, like 'consecutive integers summing to 5'. The trick of including negatives so most terms cancel pops out: -4, -3, …, 4, 5 sums to 5 and has 10 terms. Tool #1 (Diagram): a number line shows the symmetric block from -k to +k summing to 0, with a tail {k+1, …, 45} paying for the 45. Tool #5 (Pattern): once we see the smaller case, generalize — use -44 to 45 for sum 45, giving 90 terms. Tool #3 (Eliminate): the choices climb 9, 25, 45, 90, 120. Anything ≤ 45 ignores negatives. 120 is too long because the symmetric trick caps at 90.
Warm up: the integers -4, -3, …, 4, 5 sum to 5 because every pair (-k, k) cancels — that is 10 terms, far more than {5} alone.
Negatives cancel positives in pairs — the only 'survivor' is the top number.
6.NS.C.6Solve An Easier Related ProblemGeneralize: for a positive target S, the integers -(S-1) up to S sum to S because pairs cancel, giving 2S terms.
Pattern from the warm-up — sum survives, length doubles.
6.NS.C.6Look For A PatternApply to S = 45: the integers -44 up to 45 sum to 45, all pairs (-k, k) cancel, and the count is 45 - (-44) + 1 = 90.
Same pattern as the warm-up, just bigger: from -(S-1) to S gives 2S terms.
6.NS.C.6Look For A PatternConfirm it's maximal: n integers from a give n(2a + n - 1) = 90, so n must divide 90 — and 90 itself is attainable.
Length must divide 90 — and 90 itself works.
6.EE.B.7Solve An Easier Related ProblemEliminate the rest: 9 and 45 divide 90 but are smaller; 25 and 120 don't divide 90 at all — so the maximum is 90.
Among the choices, only divisors of 90 are achievable; biggest one is 90.
4.OA.B.4Eliminate PossibilitiesThis AMC 10 problem only needs Grade 6 "negatives on the number line" you already know — start at -44 and run up to 45 so every pair (-k, k) cancels and only 45 is left, giving 90 terms.