AMC 10 · 2025 · #12
Grade 7 countingPick an answer.
AMC 10 2025 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
This is a 'how many ways' count, so Tool #2 (Make a Systematic List) drives it: fix a role for each position, then multiply the independent choices. But before I can count, I must know how 'even' and 'prime' overlap among the digits 1–9, which is exactly a both/neither sorting — Tool #12 (Draw a Venn Diagram). That sorting reveals the trap digit 2, the one number sitting in the overlap. Because 2 can satisfy both conditions by itself, the count breaks into cleanly separate scenarios, so Tool #7 (Identify Subproblems) splits on whether a 2 appears and I add the pieces at the end.
Sort the digits by even and prime
Label each digit from 1 to 9 as even or not and as prime or not; the two labels together split the nine digits into four groups.
Once each digit is tagged 'even?' and 'prime?', the whole problem is just about how many digits land in each tag group.
4.OA.B.4Draw A Venn DiagramSpot the digit that fills both slots
Only 2 is both even and prime, so one 2 fills the even slot and the prime slot at the same time — that overlap forces two separate cases.
The number in the overlap of the two circles does double duty, so it behaves differently from every other digit.
2.OA.C.3Draw A Venn DiagramCase 1 — no 2 is used
Without a 2: choose the even position and its value, the prime position and its value, then fill the last two from 1 or 9 — 432 codes.
Assign one job to each position, then multiply the independent choices.
7.SP.C.8Make A Systematic ListCase 2 — exactly one 2 is used
One 2 already covers both rules, so the other three positions must each be 1 or 9: 4 places for the 2 times 8 fillings gives 32.
One lonely 2 satisfies both rules, so everything else has to be a harmless 1 or 9.
One digit satisfies both rules at once, so it has to be handled as its own case.
▸ Why?
A digit in the overlap of two conditions would be counted twice if the conditions were just added.
▸ Why?
Splitting by whether that digit is used makes the cases never overlap, so their counts simply add.
Add the two scenarios
The two cases never overlap and together cover every valid passcode, so add them: 432 plus 32 is 464.
Separate, non-overlapping cases are combined by simple addition.
7.SP.C.8Identify SubproblemsWatch for the digit that fits two rules at once (2 is even and prime) — split into 'it shows up' vs 'it doesn't', count each, and add.
- Sort the digits by even and prime
- Spot the digit that fills both slots
- Case 1 — no 2 is used
- Case 2 — exactly one 2 is used
- Add the two scenarios