Day 9: Find the Missing Digits

In base $n$ it can be proven that the sum of the digits of a number, $k$ say, is congruent to $k \bmod (n-1)$ , and that the sum of the alternating digits (from the rightmost digit) minus the sum of the remaining digits is congruent to $k \bmod (n+1)$ .

Using this result: $a+b+c+d+e+32$ is divisible by $8$ and $a+b+c+d+e+20$ is divisible by $10$ .

As $1 \leq a+b+c+d+e \leq 40$ we can consider cases until we find that only $a+b+c+d+e = 40$ satisfies the conditions. But that means that all the variables must be $8$ , meaning that the number is $\boxed{87858462888}$ as required.

We work with the congruence $\overline{a7b5c462d8e}_{9} \equiv 0 \mod 40$ .

Subtracting terms from both sides yields

$9^{10}a+9^{8}b+9^{6}c+9^{2}d+e \equiv -\overline{7050462080}_{9} \mod 40$

We know that $9^{2} = 81 \equiv 1 \mod 40$ . Applying that to the LHS, we get

$(9^{2})^{5}a+(9^{2})^{4}b+(9^{2})^{3}c+(9^{2})d+e \equiv a+b+c+d+e \mod 40$

Doing likewise to the RHS gives

$-\overline{7050462080}_{9} \equiv 9(-26)-6 \equiv 0 \mod 40$

Combining our results, the congruence $a+b+c+d+e \equiv 0 \mod 40$ implies $a = b = c = d = e = 8$ since $0 \leq a,b,c,d,e < 9$ .

Hence, our number is $\overline{a7b5c462d8e}_{9} = \boxed{\overline{87858462888}_{9}}$ .

By converting the said number to base 10 gives a7b5c462d8e (base 9) = (9^10)(a) + (9^9)(7) + (9^8)(b) + (9^7)(5) + (9^6)(c) + (9^5)(4) + (9^4)(6) + (9^3)(2) + (9^2)(d) + (9)(8) + e (base 10).

Considering modulo 8, it implies for the number in base 10 that a + b + c + d + e must be divisible by 8. It also implies for modulo 10, a + b + c + d + e is divisible by 10.

By congruences, a + b + c + d + e is divisible by lcm(8, 10) = 40. However, since a, b, c, d, and e are digits of the number in base 9, this implies that 8 >= a >= 1 and 0 <= b, c, d, e <= 8. To satisfy the congruence, it forces a = b = c = d = e = 8 implying that the number in base 9 is 87858462888.