2025 AMC 12B Problem 6

Below is the professionally curated solution for Problem 6 of the 2025 AMC 12B, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2025 AMC 12B solutions, or check the answer key.

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Concepts:place valuedivisibilitymodular arithmetic

Difficulty rating: 1390

6.

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?

77

88

1111

1414

1515

Solution:

The total in cents is 10000A+1110B+A=10001A+1110B,10000A + 1110B + A = 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}. The only digit solution with A0A \neq 0 is A=6,B=5A = 6, B = 5 (76+65=727 \cdot 6 + 6 \cdot 5 = 72), giving $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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