2025 AMC 10B Problem 8

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

A single divisibility condition on a composite modulus is best broken into coprime prime-power parts, because each part constrains a different feature of the numeral: the test for $9$ sees the digit sum and the test for $4$ sees the last two digits. Combining "$A$ even" with "$3 \mid A$" pins $A$ down in one line, which is what makes a search over $90$ digit pairs unnecessary. The modelling step is just as important: the price per shirt is an exact number of cents, so the object to test for divisibility is the integer number of cents, not the decimal number of dollars.

• Applying only the digit-sum test for 9 and stopping at the first fit A = 3, B = 4, which gives A + B = 7, choice (A).
• Testing divisibility of the decimal $655.56$ by $36$ directly instead of converting to cents first.
• Using the test for $2$ instead of the test for $4$, which leaves $B \in \{2,5,8\}$ and three surviving totals rather than one.