AMC10 2025 B

AMC10 2025 B · Q8

AMC10 2025 B · Q8. It mainly tests Divisibility & factors, Digit properties (sum of digits, divisibility tests).

Emmy says to Max, "I ordered 36 math club sweatshirts today." Max asks, "How much did each shirt cost?" Emmy responds, "I'll give you a hint. The total cost was $\$ \underline A~\underline B~\underline B.\underline B~\underline A$, where $A$ and $B$ are digits and $A \neq 0$." After a pause, Max says, "That was a good price." What is $A + B$?

Emmy 对 Max 说：“我今天订了 36 件数学俱乐部卫衣。” Max 问：“每件衬衫多少钱？” Emmy 回答：“我给你一个提示。总费用是 $\underline A~\underline B~\underline B.\underline B~\underline A$ 美元，其中 $A$ 和 $B$ 是数字且 $A \neq 0$。” 停顿后，Max 说：“这价格不错。” 求 $A + B$。

(A) 7 7

(B) 8 8

(D) 14 14

(E) 15 15

Answer

Correct choice: (C)

正确答案：（C）

Solution

The problem is essentially telling us that $\underline A~\underline B~\underline B.\underline B~\underline A$ is divisible by $36$, so it is divisible by $9$. Then, by the divisibility by $9$ condition, \[A+B+B+B+A=2A+3B\equiv 0\pmod{9}\] This means that $A$ must be divisible by $3$, or otherwise $2A+3B$ would not be divisible by $3$ and thus $9$. Since $A\ne0$, we must have one of $A=3,6,9$. But $A$ must be even or else the entire number would not be even and thus would not be divisible by $4$. Hence $A=6$. Then, $2\cdot 6+3B\equiv 0\pmod{9}$, so $4+B\equiv 0\pmod{3}$, and thus $B\equiv 2\pmod{3}$. This yields $B=2,5,8$, of which $B=5$ is the only number that allows $\underline A~\underline B~\underline B.\underline B~\underline A$ to be divisible by $4$. Thus the answer is $6+5=\boxed{\textbf{(C)}~11}$.

问题本质上是告诉我们 $\underline A~\underline B~\underline B.\underline B~\underline A$ 能被 36 整除，因此能被 9 整除。然后，根据能被 9 整除的条件， \[A+B+B+B+A=2A+3B\equiv 0\pmod{9}\] 这意味着 $A$ 必须能被 3 整除，否则 $2A+3B$ 不能被 3 整除从而不能被 9 整除。由于 $A\ne0$，必须有 $A=3,6,9$ 中的一个。但 $A$ 必须是偶数，否则整个数不是偶数从而不能被 4 整除。因此 $A=6$。然后，$2\cdot 6+3B\equiv 0\pmod{9}$，所以 $4+B\equiv 0\pmod{3}$，从而 $B\equiv 2\pmod{3}$。这给出 $B=2,5,8$，其中只有 $B=5$ 使 $\underline A~\underline B~\underline B.\underline B~\underline A$ 能被 4 整除。因此答案是 $6+5=\boxed{\textbf{(C)}~11}$。

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