2025 AMC 10B Problem 8

Below is the professionally curated solution for Problem 8 of the 2025 AMC 10B, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2025 AMC 10B solutions, or check the answer key.

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Concepts:divisibilitymodular arithmeticplace value

Difficulty rating: 1350

8.

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 \ne 0." After a pause, Max says, "That was a good price." What is A+B?A + B?

77

88

1111

1414

1515

Solution:

In cents the total is 10000A+1000B+100B+10B+A=10001A+1110B.10000A + 1000B + 100B + 10B + A = 10001A + 1110B. Split evenly among 3636 shirts, so it's divisible by 36.36. Now 100012910001 \equiv 29 and 111030(mod36),1110 \equiv 30 \pmod{36}, so we need 29A+30B0,29A + 30B \equiv 0, which reduces to 7A+6B0(mod36).7A + 6B \equiv 0 \pmod{36}. The only digit solution with A0A \ne 0 is A=6,B=5,A = 6, B = 5, since 76+65=72.7 \cdot 6 + 6 \cdot 5 = 72. That's $655.56,\$655.56, or $18.21\$18.21 a shirt, so A+B=11.A + B = 11. Therefore, the answer is C.

Problem 8 in Other Years