2025 AMC 10B Problem 22

The sample space. Seven digits sum to at most $7 \cdot 9 = 63$, so a digit sum of $61$ means the digits are all $9$ except for a total shortfall of $2$. There are exactly two ways to distribute that shortfall:

• one digit is $7$ and the other six are $9$: $7$ numbers, one for each position;
• two digits are $8$ and the other five are $9$: $\binom{7}{2} = 21$ numbers.

Every digit is at least $7$, so the leading digit is never $0$ and all $28$ strings are genuine seven-digit numbers. The condition leaves $28$ equally likely numbers.

The divisibility test. A number is divisible by $11$ exactly when its alternating digit sum
$$A = d_1 - d_2 + d_3 - d_4 + d_5 - d_6 + d_7$$
is a multiple of $11$. Four positions ($1,3,5,7$) carry a plus sign and three ($2,4,6$) carry a minus sign, so if every digit were $9$ then
$$A = 4\cdot 9 - 3 \cdot 9 = 9 .$$

Lowering a digit in a plus position decreases $A$ by the same amount; lowering one in a minus position increases $A$. The total shortfall is $2$, so
$$7 \le A \le 11,$$
and the only multiple of $11$ in that range is $11$. So $A = 11$, which happens exactly when the whole shortfall of $2$ sits in minus (even) positions.

Counting.

• One $7$. It must occupy one of the three even positions: $3$ numbers.
• Two $8$s. Both must occupy even positions: $\binom{3}{2} = 3$ numbers.

That is $6$ numbers out of $28$, so the probability is
$$\frac{6}{28} = \frac{3}{14}.$$

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

Conditioning on a digit sum that sits right at the top of its range collapses an enormous sample space to a list short enough to handle by hand — the useful reformulation is "shortfall from all nines", not "digits summing to 61". The $11$-test is linear in the digits, so instead of testing numbers one at a time, compute the alternating sum of the all-nines baseline and track how each shortfall moves it. That also produces the bound $7 \le A \le 11$ for free, which is what rules out every multiple of $11$ except one and removes all remaining casework.

• Noticing only one of the two ways to fall 2 short of 63; keeping just the numbers with a single 7 gives 3 out of 7, which is (E).
• Requiring the alternating sum to be $0$ rather than any multiple of $11$, which rejects all $28$ numbers.
• Miscounting the plus and minus positions: with seven digits there are four plus positions and three minus positions, so the baseline is $9$, not $0$.