Tessellate S.T.E.M.S. (2019) - Mathematics - School - Set 1 - Objective Problem 4

We claim that the desired number is $598 999 \cdots 99989888$ (with $217$ times $9$ in between).

It is clear that the number has more than $224$ digits, because $2016 ÷ 9 = 224$ . Now, to minimize the number we need to find a number with $225$ digits which is a multiple of $2016$ . We write

$N = \overline{a_{224}a_{223} \cdots a_4a_3a_2a_1a_0}$

Since $2016 = 2^5 . 3^2 . 7$ , it is necessary and sufficient that the number $k\overline{ a_4a_3a_2a_1a_0}$ is a multiple of $32$ and the number $N$ is a multiple of $7$ , since we already have $N \equiv s(N) = \sum_{i=0}^{224} a_i = 2016 \equiv 0 \pmod{9}$ .

First, we will analyze $s(k) = \sum_{j=0}^4 a_j$ . The digit $a_0$ is even and is least equal to $8$ . Thus we have the following cases:

• If $s(k) = 44$ , then $k$ is $99998$ and all its permutations.

• If $s(k) = 43$ , then $k$ is $99988$ and all its permutations.

• If $s(k) = 42$ , then $k$ is $99888, 99978$ , or $99996$ and all its permutations.

• If $s(k) = 41$ , then $k$ is $98888, 99878$ , or $99968$ and all its permutations.

We discard most of the numbers since the ones terminating in $7, 9, 78, 86, 98, 788, 988, 996$ or $8888$ are not multiples of $32$ . The only multiples of $32$ in all of these cases are $99968$ and $89888$ , and for both of these, $s(k) = 41$ .

If $a_{224} \leq 3$ then $\sum_{i=0}^{224} a_i \leq 3+219 . 9+\sum_{j=0}^4 a_j = 1974+s(k) \implies s(k) \geq 42$ , a contradiction, as $s(k) \leq 41$ .

If $a_{224} = 4$ , then $2016 = \sum_{i=0}^{224} a_i \leq 4 + 219 . 9 + s(k) = 1975 + s(k)$ , then $s(k) \geq 41$ , and $s(k) = 41$ , whence $N$ can terminate in only two ways, but since $99968 \equiv 89888 \equiv 1 \pmod{ 7}$ , to minimize we choose $89888$ .

$N = 4999 \cdots 99989888$ ( $219$ times $9$ in between) $= 4 . 10^{224} + (10^{219} - 1) . 10^5 + 89888 \equiv 4 . 2 + (6 - 1) . 5 + 1 = 6 \not\equiv 0 \pmod{ 7}$ .

If $a_{224} = 5$ , then $2016 = \sum_{i=0}^{224} a_i \leq 5 + 219 . 9 + s(k) = 1976 + s(k)$ , whence $s(k) \geq 40$ . To minimize we must have $s(k) = 41$ and as in the previous case, $N$ terminates in $89888$ .

$N = 5 999 \cdots 8 \cdots 99989888$ ( $219$ times $9$ in total) $= 5 . 10^{224} + (10^{219} - 1) .10^5 - 10^m + 89888$ where $5 \leq m \leq 223$ .

$N \equiv 5 . 2 + (6 - 1) . 5 - 10^m + 1 \equiv 1 - 10^m \equiv 0 \pmod{7}$ . Then $10^m \equiv 1 \pmod{7} \iff m \equiv 0 \pmod{ 6}$ . To minimize we choose the highest value for $m$ , namely, $m = 222$ .