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2025 AMC 12B Problem 6

Problem 6 of 25EasierNumber Theory

Emmy says to Max, “I ordered 3636 math club sweatshirts today.” Max asks, “How much did each shirt cost?” Emmy responds, “I’ll give you a hint. The total cost was $ABB.BA,\$\underline{A}\,\underline{B}\,\underline{B}.\underline{B}\,\underline{A}, where AA and BB are digits and A0.A \neq 0.” After a pause, Max says, “That was a good price.” What is A+B?A + B?

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Solution

The total in cents is 10000A+1110B+A10000A + 1110B + A =10001A+1110B,= 10001A + 1110B, which must be a multiple of 36.36. Since 100012910001 \equiv 29 and 111030(mod36),1110 \equiv 30 \pmod{36}, the condition is 29A+30B0,29A + 30B \equiv 0, i.e. 7A+6B0(mod36).7A + 6B \equiv 0 \pmod{36}. Reducing once more modulo 66 gives A0(mod6),A\equiv0\pmod6, so the only possible nonzero digit is A=6.A=6. Then 42+6B42+6B is divisible by 3636 only for the digit B=5.B=5. Indeed $655.56=36×$18.21,\$655.56 = 36 \times \$18.21, so A+B=11.A+B=11. Thus, the correct answer is C.

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Concepts: place value · divisibility · modular arithmetic

Problem text and solution from the LIVE past-contest archive. See also the AoPS wiki page for community solutions.