2025 AMC 10B Problem 8

P1-10~4 minPrint

Divisibility & Factors Modular Arithmetic Number Properties Split into exhaustive cases and handle each Set up the equation/formula and compute; no special trick needed

Emmy says to Max, "I ordered $36$ math club sweatshirts today." Max asks, "How much did each shirt cost?" Emmy responds, "I'll give you a hint. The total cost was $\$ \underline A~\underline B~\underline B.\underline B~\underline A$ , where $A$ and $B$ are digits and $A \neq 0$ ." After a pause, Max says, "That was a good price." What is $A + B$ ?

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Key insight

The total in cents, ABBBA, is a multiple of 36. The test for 4 makes A even; the test for 9 makes A a multiple of 3.

Solution

Work in cents. Each shirt costs a whole number of cents. So the total in cents, the five-digit number $\underline A\,\underline B\,\underline B\,\underline B\,\underline A$ , must split evenly into $36$ equal parts. In other words, it is a multiple of $36$ .

Split 36 into two easy tests. A number is a multiple of $36$ exactly when it is a multiple of both $4$ and $9$ . Each has a simple digit rule.

Test for 9: the digit sum. The digits add to $A + B + B + B + A = 2A + 3B$ , and this must be a multiple of $9$ . In particular it must be a multiple of $3$ . The part $3B$ already is, so $2A$ must be too, and that means $A$ is a multiple of $3$ : $A$ is $3$ , $6$ or $9$ .

Test for 4: the last two digits. A multiple of $4$ is even, so its last digit, $A$ , must be even. (The full test, that the last two digits $\underline B\,\underline A$ form a multiple of $4$ , comes back in step 6.)

Find $A$ . The only digit that is even and a multiple of $3$ (and not $0$ ) is $A = 6$ .

Find $B$ . The digit sum is $12 + 3B$ . Try $B = 0, 1, \ldots, 9$ : it is a multiple of $9$ only for $B = 2, 5, 8$ (sums $18, 27, 36$ ). The last two digits are $26$ , $56$ or $86$ , and only $56$ is a multiple of $4$ . So $B = 5$ .

Check. The total is $\$655.56$ , which is $65556$ cents, and $65556 \div 36 = 1821$ . Each shirt cost $\$18.21$ , a good price.

So $A + B = 6 + 5 = 11$ .

The answer is $\boxed{\textbf{(C)}\ 11}$ .

Why this works

A test for a big number like $36$ is easiest when you break it into smaller numbers with no common factor, here $4$ and $9$ . Each smaller test looks at a different part of the number: the test for $9$ uses the digit sum, and the test for $4$ uses the last two digits. Put together, they pin down $A$ right away. The key first step is to count in cents, so the total is a whole number you can test.

Concise solution

A shorter write-up, the way contest solutions are usually written. It may take a different route from the solution above.

Each sweatshirt costs a whole number of cents, so the total measured in cents, which is the five-digit number $\underline{A}\,\underline{B}\,\underline{B}\,\underline{B}\,\underline{A}$ , must be divisible by $36$ . Since $36 = 4 \cdot 9$ with $\gcd(4,9) = 1$ , test both factors.

Divisibility by 9. The digit sum is $A + B + B + B + A = 2A + 3B$ , so $2A + 3B \equiv 0 \pmod 9$ . In particular $3 \mid 2A$ , and since $3 \nmid 2$ this forces $3 \mid A$ .

Divisibility by 4. Only the last two digits matter: $4 \mid 10B + A$ . As $10B$ is even, $A$ must be even.

So $A$ is a nonzero digit that is both even and a multiple of $3$ : the only possibility is $A = 6$ .

Now return to the two tests with $A = 6$ . From $12 + 3B \equiv 0 \pmod 9$ we get $3B \equiv 6 \pmod 9$ , so $B \equiv 2 \pmod 3$ and $B \in \{2, 5, 8\}$ . From $4 \mid 10B + 6$ , that is $2B + 2 \equiv 0 \pmod 4$ , we get $B$ odd. Only $B = 5$ survives.

The total is $\$655.56$ , or $65556$ cents, and $65556 = 36 \cdot 1821$ , so each shirt cost $\$18.21$ — a good price, and the fact that Max can name it confirms the digits are unique. Hence $A + B = 6 + 5 = 11$ .

The answer is $\boxed{\textbf{(C)}\ 11}$ .

Review in the AoPS books

Introduction to Number Theory, Chapter 1: Integers: The Basics (§1.4, §1.5)
Introduction to Number Theory, Chapter 12: Introduction to Modular Arithmetic
Introduction to Counting & Probability, Chapter 2: Basic Counting Techniques (§2.2)
Introduction to Number Theory, Chapter 5: Divisor Problems (§5.2, §5.3)