๐Ÿ”ฅ New! 2025 AMC 10 B

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Problem 22 Easy

A seven-digit positive integer is chosen at random. What is the probability that the number is divisible by 11, given that the sum of its digits is 61?

  • A.

    \frac { 3 } { 1 4 }

  • B.

    \frac { 3 } { 1 1 }

  • C.

    \frac { 2 } { 7 }

  • D.

    \frac { 4 } { 1 1 }

  • E.

    \frac { 3 } { 7 }

Answer:A
Solution

Given (x_1 \cdots x_7)_{10} with \sum_{i=1}^{7} x_i = 61 and 0 \leq x_i \leq 9.

Let y_i = 9 - x_i, then \sum_{i=1}^{7} y_i = 2 and 0 \leq y_i \leq 9.

This gives \sum_{i=1}^{7}(y_i + 1) = 9, so there are \binom{8}{6} = \binom{8}{2} = 28 numbers.

From \sum_{i=1}^{7} y_i = 2, we have y_i \leq 2 and x_i \geq 7.

Let a = x_2 + x_4 + x_6, then x_1 + x_3 + x_5 + x_7 = 61 - a.

For 11 \mid (x_1 \cdots x_7)_{10}, we need 61 - a \equiv a \pmod{11}, which gives a \equiv 3 \pmod{11}.

Since a \geq 7 \times 3 = 21 and a \leq 3 \times 9 = 27, we must have a = 11 \times 2 + 3 = 25.

Then 61 - a = 36, so x_1 = x_3 = x_5 = x_7 = 9 and y_1 = y_3 = y_5 = y_7 = 0.

For y_2 + y_4 + y_6 = 2 with y_2, y_4, y_6 \geq 0 and y_i \in \mathbb{Z}:

(y_2+1) + (y_4+1) + (y_6+1) = 5 has \binom{5-1}{3-1} = 6 solutions.

Therefore, there are 6 numbers in total, with probability = \frac{6}{28} = \frac{3}{14}.

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