Competition · AMC preparation · step 4 of 4
AMC 10 · 2023A · #12
Grade 4 number-theoryPick an answer.
AMC 10 2023 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Two conditions act on different digits, so Tool #7 (Identify Subproblems) splits the work: first use the reversal-divisibility-by-5 rule to pin down the hundreds digit of N, then use divisibility by 7 to count how many of the resulting three-digit numbers survive. The divisibility-by-5 subproblem collapses to a single hundreds digit because a ≠ 0. Tool #2 (Systematic List) underpins step two — listing multiples of 7 in the locked-down range, then using Tool #3 (Eliminate) to cross-check the final count against the choices.
Find the leading digit
Write N as abc; reversed it's cba, whose units digit is a. Divisible by 5 forces that digit to 0 or 5, so a ∈ {0, 5}.
The divisibility-by-5 rule (units digit 0 or 5) is the very first divisibility rule kids meet — pure Grade 4 number-sense.
Divisibility by five is decided entirely by the last digit.
▸ Why?
Every higher place value is already a multiple of five, so it leaves the remainder untouched.
▸ Why?
A number is its digits weighted by their places, so the last place can be examined alone.
Narrow the range
A three-digit number's hundreds digit can't be 0, so a must be 5 — every valid N lies in 500 to 599.
The Grade 2 place-value rule "hundreds digit can't be 0" knocks a = 0 out and pins the whole search to the 500s.
2.NBT.A.1Identify SubproblemsFind the first multiple of 7
Divide to find the first multiple of 7 at or above 500: 500 ÷ 7 = 71 r 3, so it's 7 × 72 = 504.
Dividing with remainder (Grade 4) tells you exactly where the next clean multiple lands — no guess-and-check needed.
4.NBT.B.6Identify SubproblemsFind the last multiple of 7
Same for the top end: 599 ÷ 7 = 85 r 4, so the largest multiple of 7 at or below 599 is 7 × 85 = 595.
Same divide-with-remainder trick from the upper end — the last clean multiple is one step before 599.
4.NBT.B.6Identify SubproblemsList the surviving numbers
List the survivors in order: 7 × 72, 7 × 73, …, 7 × 85 — an arithmetic run indexed by 72 to 85.
Following the rule "add 7 each time, stop at 595" generates the list in order — Grade 4 pattern generation.
4.OA.C.5Make A Systematic ListCount the list
Count consecutive integers from 72 to 85: 85 - 72 + 1 = 14.
"Last minus first plus one" is the basic Grade 3 counting move for consecutive integers.
3.OA.D.8Make A Systematic ListCheck one candidate
Cross-check 504: reversal 405 ends in 5 (÷5 ✓) and 504 = 7 × 72 (÷7 ✓) — both hold, matching choice (B).
Verifying one item from the list against both conditions confirms the whole list is on the right track.
4.OA.B.4Eliminate PossibilitiesTwo divisibility conditions split into two small steps: the "reversed is a multiple of 5" rule pins down the leading digit (it has to be 5), and then you just count the multiples of 7 between 500 and 599 — fourteen of them. Grade 4 divisibility rules carry the whole problem.
- Find the leading digit
- Narrow the range
- Find the first multiple of 7
- Find the last multiple of 7
- List the surviving numbers
- Count the list
- Check one candidate
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