Find any 100 digit number with non-zero digits divisible by the sum of its digits.

ricee Apr 28, 2021

#1**+2 **

The easiest divisibility rule to work with (at least for me) is the divisibility rule of 2^n, where n is some positive integer. That rule says that if the last n digits of any number is divisible by 2^n, then the whole number is.

Since we need to find a number bigger than 100, n must be at least 7, since 2^7 = 128 > 100.

The smallest 7 digit multiple of 128 with no digits equal to zero is 1111168 because 1111111 is the smallest 7 digit number with no digits equal to zero, and \(\lfloor{\frac{1111111}{128}}\rfloor\cdot128=1111168\)

The rest of the 93 digits have to sum up to 128-1-1-1-1-1-6-8=109. Obviously, there is a lot of ways to make 93 one-digit numbers sum to 109, but the problem only asks for 1 example, so you could just use ninety 1's, two 5's, and one 9, for the rest of the digits of the number.

The full number would look like this: \(\boxed{1111111111111111111111111111111111111111111111111111111111111111111111111111111111111111115591111168}\)

(you could check in WA or some other calculator to see if it's correct)

textot Apr 28, 2021