AMC 10 · 2025 · #8
Grade 6 number-theoryPick an answer.
AMC 10 2025 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
The digits A and B can only take a handful of values, so the smart move is to turn each fact into a divisibility test and cross off every pair that fails. First read the price as a whole number of cents so 'divisible by 36' means something exact, then use the rules for 9 and 4 to squeeze the candidates down to a single pair.
Read the price as whole cents
In cents the total is the five-digit number ABBBA, and 36 shirts splitting it evenly means 36 divides that number.
Money in cents is just a plain whole number, so 'splits evenly among 36' becomes a clean divisibility question.
5.NBT.A.3Analyze The UnitsSplit 36 into 4 and 9
Since 36 = 4 × 9 and 4 and 9 share no factor, ABBBA must pass the rule for 4 and the rule for 9 at once.
Breaking 36 into 4 and 9 lets you use two easy digit rules instead of dividing a big number.
Breaking the divisor into two coprime pieces lets you use two easy digit rules instead of one hard division.
▸ Why?
Two pieces with different prime recipes share nothing, so meeting both rules is meeting the whole one.
▸ Why?
Each of those rules reads off the digits directly, since the place values line up with them.
Apply the rule for 9
The digit sum is 2A + 3B; 3B is already a multiple of 3, so A must be one too, leaving A = 3, 6, or 9.
Since 3B is already a multiple of 3, A has to carry the rest of the load and be a multiple of 3 too.
6.EE.B.6Introduce A VariableUse divisibility by 4 to pin A
Divisible by 4 means even, and the last digit is A, so of 3, 6, 9 only A = 6 survives.
Divisible by 4 forces an even last digit, and 6 is the only even option left for A.
4.OA.B.4Eliminate PossibilitiesNarrow B with the rule for 9
With A = 6 the digit sum 12 + 3B is a multiple of 9 only when B leaves remainder 2 mod 3: B = 2, 5, or 8.
Locking A at 6 turns the rule for 9 into a short list of possible B values.
6.EE.B.6Introduce A VariableFinish with the rule for 4
The last two digits make 10B + 6, and of 26, 56, 86 only 56 is a multiple of 4, so B = 5 and A + B = 11 — choice (C).
The last two digits '56' are the only ones divisible by 4, which uniquely fixes B.
4.OA.B.4Eliminate PossibilitiesTo check divisibility by 36, split it into 4 and 9 and use the two easy digit rules to cross off every impossible pair until one survives.
- Read the price as whole cents
- Split 36 into 4 and 9
- Apply the rule for 9
- Use divisibility by 4 to pin A
- Narrow B with the rule for 9
- Finish with the rule for 4