AMC10 2025 B

Let the number be represented as $\overline{a_1a_2a_3a_4a_5a_6a_7}.$ We first seek to find the number of seven-digit integers with $a_1+a_2+a_3+a_4+a_5+a_6+a_7 = 61.$ We would love to do stars and bars here, but we have the condition that $0 \leq a_n \leq 9.$ Therefore, we define $a_n' = 9-a_n \implies a_n = 9-a_n'.$ This is a bijective operation, so the total number of integers doesn't change when we use these digits instead. Substituting in gives that ∑k=17(9−ak′)=61,a1′+a2′+a3′+a4′+a5′+a6′+a7′=2. Since the $a_n'$ are nonnegative integers and the condition $a_n' \leq 9$ doesn't matter with a sum less than 9, we can do stars and bars to figure out the number of permutations of digits that satisfy the constraint: ${{2+6}\choose {2}} = {8\choose 2} = 28.$ Now, we need to find how many of these integers are divisible by $11.$ We can use the divisibility by $11$ rule, which says that $(a_1+a_3+a_5+a_7) - (a_2+a_4+a_6) = 11k$ for some integer $k.$ Notice that $k \geq 1$ because, otherwise, $a_2+a_4+a_6 > 27$ which is the maximum sum of three digits. In addition, $k \leq 1$ because, otherwise $a_1+a_3+a_5+a_7 > 36$ which is the maximum sum of four digits. Therefore $k = 1.$ Then we have that $(a_1+a_3+a_5+a_7) - (a_2+a_4+a_6) = 11,$ so $a_1+a_3+a_5+a_7 = 36$ and $a_2+a_4+a_6 = 25.$ The first equation can only result in $a_1=a_3=a_5=a_7 = 9.$ The second equation can be rewritten in a similar manner as before to get that (9−a2′)+(9−a4′)+(9−a6′)=25,a2′+a4′+a6′=2. Again, with stars and bars, there exists ${{2+2}\choose {2}} = {4\choose 2} = 6$ ordered triples of $(a_2, a_4, a_6)$. Therefore, our answer is \[\frac{6}{28} = \boxed{\text{(A) } \frac{3}{14}}.\]