2025 AMC 10B Problem 8

1. 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 $36$.

1. 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.

1. 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 $9$.

1. 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.)

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

1. 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 $B = 5$.

1. 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 $\boxed{\textbf{(C)}\ 11}$.

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.

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 $\boxed{\textbf{(C)}\ 11}$.