2025 AMC 10B Problem 22

Step 1: list the possible numbers. Seven 9s add to $63$. A digit sum of $61$ is $2$ less, so the digits are all 9s except for a total "shortfall" of $2$. That happens in two ways:
- one digit is $7$ and the rest are $9$: the $7$ can be in any of $7$ places, so $7$ numbers;
- two digits are $8$ and the rest are $9$: choose $2$ of the $7$ places, $\binom72 = 21$ numbers.

That is $28$ numbers, all equally likely. (None starts with $0$, since every digit is at least $7$.)

Step 2: the test for 11. Number the places $1$ to $7$ from the left. A number is a multiple of $11$ exactly when
$$(\text{digits in places } 1, 3, 5, 7) - (\text{digits in places } 2, 4, 6)$$
is a multiple of $11$ (this can be $0$ or negative too).

Step 3: start from 9999999. For all 9s, this difference is $4 \times 9 - 3 \times 9 = 9$.
- Lowering a digit in an odd place (1, 3, 5, 7) makes the difference smaller by that amount.
- Lowering a digit in an even place (2, 4, 6) makes the difference bigger by that amount.

The total lowering is $2$, so the difference is somewhere from $9 - 2 = 7$ to $9 + 2 = 11$. The only multiple of $11$ in that range is $11$. So we need the difference to be exactly $11$, which means all of the shortfall must be in even places.

Step 4: count the good numbers.
- One $7$: it must be in place $2$, $4$ or $6$. That gives $3$ numbers.
- Two $8$s: both must be in even places, $\binom32 = 3$ ways. That gives $3$ numbers.

So $6$ of the $28$ numbers work, and the probability is
$$\frac{6}{28} = \frac{3}{14}.$$

The answer is $\boxed{\textbf{(A)}\ \frac{3}{14}}$.

A digit sum close to the largest possible value leaves very few numbers, so thinking of "how far short of all 9s" makes the list tiny. The test for $11$ adds and subtracts digits, so instead of testing $28$ numbers one by one, start from $9999999$ and track how each lowered digit changes the result. The range $7$ to $11$ leaves only one possible multiple of $11$, which settles everything.