AMC 10 · 2021 · #16
Grade 4 number-theoryPick an answer.
AMC 10 2021 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Tool #9 (Easier Problem) — first cut the candidate pool by using the divisibility rules. "Divisible by 5" forces the last digit to be 0 or 5, but 0 as the last digit of an uphill integer would require negative previous digits — impossible. So the last digit must be 5, and every other digit comes from {1, 2, 3, 4}. That shrinks the universe from "all uphill integers" to "subsets of {1, 2, 3, 4} followed by 5." Tool #2 (Systematic List) then enumerates every such subset in order of size (empty, singletons, pairs, triples, the full set) and Tool #5 (Pattern — divisibility-by-3 rule) filters: keep only the ones whose digit sum is a multiple of 3. Count what survives.
15 = 3 × 5, so divisible by 15 means divisible by both 3 and 5. An uphill integer can't end in 0, so its last digit must be 5.
Split the hard rule "divisible by 15" into two simpler rules, then use the easier one (÷ 5) to lock the last digit.
4.OA.B.4Solve An Easier Related ProblemEnding in 5 with increasing digits, the earlier ones form a subset of {1, 2, 3, 4} in order — pick a subset, then append 5.
Pick the digits going up to 5 — order is forced by "uphill", so just pick the set.
4.OA.B.4Make A Systematic ListSize 0 (just "5"): digit sum = 5. Not a multiple of 3. Skip.
5 alone is divisible by 5 but not by 3, so not by 15.
4.OA.B.4Look For A PatternSize 1 (one digit before 5): 15, 25, 35, 45 have sums 6, 7, 8, 9; multiples of 3 are 15 and 45 — 2 found.
Sum of digits is 3 + (extra digit), so the extra digit must be a multiple of 3 — namely 1 won't work, but 1 gives sum 6 — wait, just check directly: 1 and 4 work.
4.OA.B.4Make A Systematic ListSize 2 (two digits before 5): the 6 subsets give sums 8, 9, 10, 10, 11, 12; only 135 and 345 clear 3 — 2 found.
List all pairs from {1,2,3,4} in order, append 5, check digit sums.
4.OA.B.4Make A Systematic ListSize 3 (three digits before 5): 1235, 1245, 1345, 2345 have sums 11, 12, 13, 14; only 1245 works — 1 found.
Same drill: list, sum digits, check ÷ 3.
4.OA.B.4Make A Systematic ListSize 4 (all of 1, 2, 3, 4): 12345 has digit sum 15, divisible by 3 — 1 more. Total 2 + 2 + 1 + 1 = 6, choice (C).
Add up the counts from each size — six uphill multiples of 15 in all.
4.OA.B.4Make A Systematic ListThis AMC 10 problem only needs Grade 4 divisibility rules and a careful list you already know! Divisible by 15 means divisible by 3 AND divisible by 5, so the last digit is 5, the other digits come from {1,2,3,4}, and the digit sum must be a multiple of 3. Listing every subset of {1,2,3,4} gives exactly 6 uphill multiples of 15.