AMC 10 · 2023 · #16
Grade 8 arithmeticPick an answer.
AMC 10 2023 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Once you notice that picking a set of digits and writing them in order is the WHOLE choice, the counting collapses: U = (subsets of {1,…,9} of size ≥ 2), D = same PLUS the extra downnos that contain 0. Tool #9 (small cases) builds the bijection insight, Tool #2 (systematic list) checks the small case by hand, and Tool #16 (focus on the difference D - U) reduces |U-D| to a single subset count.
Warm up on 2-digit cases: upnos = C(9,2)=36, downnos = C(10,2)=45; the gap 9 is exactly the downnos ending in 0.
Trying the smallest case (2 digits) reveals the whole pattern: the only difference between upnos and downnos is the extra digit 0 available to downnos.
4.OA.B.4Solve An Easier Related ProblemGeneralize: each number is fixed by its digit-set, so U counts size-≥2 subsets of {1..9} and D counts size-≥2 subsets of {0..9}.
Once digits are picked, the order is forced — so counting upnos/downnos becomes counting subsets, a pure combinations question.
7.SP.C.8Make A Systematic List0-free downno sets match the upno sets one-for-one, so the whole difference is D - U = the downnos whose set contains 0.
Pair each upno's digit-set with the same downno's digit-set; the leftover downnos are exactly the ones containing 0. Counting the leftover is the whole job.
7.SP.C.8Count The ComplementCount 0-containing sets: any non-empty subset of {1..9} plus 0, giving 2⁹ - 1 = 511, so |U - D| = 511.
Each of the 9 non-zero digits is independently "in" or "out", giving 2⁹ subsets; subtract the empty one because we need at least one partner for 0.
8.EE.A.1Count The ComplementAn upno is just a set of digits from {1,…,9} written low-to-high; a downno can also use 0 (placed at the end). The two counts match perfectly EXCEPT for downnos that include 0, and those are in one-to-one correspondence with non-empty subsets of {1,…,9}. So |U-D| = 2⁹ - 1 = (E) 511.