Competition · AMC preparation · step 4 of 4
AMC 8 · 2000 · #11
Grade 4 number-theoryPick an answer.
AMC 8 2000 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
There are only 39 numbers in the range, so a brute-force check is plausible — but smarter is Tool #2 (Make a Systematic List) grouped by the units digit u. Each u from 1 to 9 gives at most 4 candidates (1u, 2u, 3u, 4u), and the question "is N divisible by u?" is identical for the whole group. Tool #3 (Eliminate Possibilities) then knocks out non-multiples within each group with a single divisibility rule. Splitting by units digit turns one 39-item check into nine tiny checks of at most 4 each — much easier to count without errors.
Split into cases by units digit
Fix the units digit u (skip 0). For each u the candidates are always 1u, 2u, 3u, 4u — four numbers — and we ask which are multiples of u.
Grade 4 place value: a two-digit number is (tens digit) × 10 + (units digit). Fixing u leaves only the tens digit to vary.
4.NBT.A.2Make A Systematic ListCount the easy cases
Easy digits pass all four: u=1 (every number is divisible by 1), u=2 (ends in 2, even), u=5 (ends in 5) — that is 12 winners.
Three digits (1, 2, 5) come free because divisibility by them is built into how the number ends — Grade 4 divisibility rules.
For units digits 1, 2, and 5, every candidate in the range is divisible by its own units digit, whatever the tens digit happens to be.
▸ Why?
A number whose units digit is 1 is divisible by 1, because one whole copy of any number is just that number itself.
▸ Why?
A number whose units digit is 2 or 5 splits into a tens part plus that final digit, and both pieces are already multiples of the final digit, so the whole number is a multiple of it too — and this stays true for any tens digit.
▸ Why?
Reading a two-digit number as (its tens digit) tens plus (its units digit) ones is just naming its place values.
▸ Why?
Ten equals two 5s and also five 2s, so any whole number of tens is a multiple of both 2 and 5, because it is built from equal groups of that digit.
▸ Why?
The final digit is one copy of itself, so it is a multiple of itself, and the shared digit can be pulled out of both pieces at once, showing their sum is still a multiple of that digit.
Count the medium cases
Medium cases: u=4 keeps 24 and 44; u=3 only 33; u=6 only 36; u=8 only 48 — that adds 5 more.
Just multiplication-table recall: scan the multiples of u up to about 50 and keep the ones whose units digit really is u.
3.OA.C.7Eliminate PossibilitiesCheck the empty cases
Hard but empty: multiples of 7 and of 9 never end in 7 or 9 in range — the next would be 77 and 99 — so both give 0.
A multiple of u that also ends in u must come from u · k where k ≡ 1 (mod 10/gcd(u,10)). For u = 7, 9 the next such k after 1 is 11, giving 77, 99 — both outside [11, 49].
4.OA.B.4Eliminate PossibilitiesAdd all nine counts
Add every case: 4+4+1+2+4+1+0+1+0 = 17, choice (C).
The systematic list is exhaustive and non-overlapping (every number has exactly one units digit), so adding the case counts gives the total without double-counting.
3.NBT.A.2Make A Systematic ListSort the 39 numbers by their last digit and check each group with one divisibility rule. Three easy digits (1, 2, 5) give 12 winners on their own, and the rest contribute 5 more — total 17, answer (C).
- Split into cases by units digit
- Count the easy cases
- Count the medium cases
- Check the empty cases
- Add all nine counts
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