Tessellate S.T.E.M.S. (2019) - Computer Science - School - Set 3 - Objective Problem 1

$S(x)$ , for a natural number $x$ is the digital sum of $x$ . First, we find the minimum number digits $N$ can have. It would be $12$ , for if we have $11$ digits, then $S(N) \leq 9\times 11 < 100$ .

$a_i$ , for $i \in \{0,1,\dots,9\}$ , is the number times a digit $i$ is repeated in $N$ . then the following equations hold true.

$\sum_{i=0}^{9}ia_i=100$

$a_1+2a_2+3a_3+4a_4-4a_5-3a_6-2a_7-a_8=10$

for the second equation, you need to pay attention to the digital difference, made when a single-digit number is multiplied by two. For example, $2$ becomes $4$ , after being multiplied by $2$ and the difference (made to the digital sum) would be $+2$ , regardless of the position of $2$ in the number or its preceding and succeeding digits. for $8$ , the difference would be $-1$ .

Now, we subtract the second equation from the first equation to get

$9(a_5+a_6+a_7+a_8+a_9)=90 \implies (a_5+a_6+a_7+a_8+a_9)=10$ . So, there should be a total of $10$ digits from digits $5,6,7,8,9$ .

$N$ cannot have $12$ digits. because even if we have $10$ $9$ s and the remaining $2$ digits are $4$ , $S(N)=98<100$ . So we may try $13$ digits. one can easily find candidates like $4448899999999$ . from there, you can modify the number, so it becomes smaller. Because of the symmetry, one can take $1$ off of one the first three digits and add it to one of the last ten digits, as long as we do not hit $0$ for any of the last ten digits. after the process, we are forced to stop at $2449999999999$ , which is the final answer.