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

 
 Apr 28, 2021
 #1
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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)

 
 Apr 28, 2021
 #2
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0

thanks for help smiley

 
ricee  Apr 28, 2021

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