Wednesday, June 20th

Problem
Find all positive integers that are 700 times the sum of its digits.

Solution
This statement is equivalent to the 700 replaced by 7. If $d$ is the number of digits, then the maximum value of 7 times the sum of digits is $63d$ and the minimum value of the number is $10^{d-1}$, hence the maximum value of $d$ is 3. Let $n=100a+10b+c$, where $a,b,c$ are digits, then

$100a+10b+c=7a+7b+7c$

and hence $93a+3b=6c$. 


Therefore $a=0$ and $b=2c$. The only posibilities for $b$ and $c$ are $(2,1)$, $(4,2)$, $(6,3)$, and $(8,4)$, thus the only numbers that satisfy the problem are $m=2100, 4200,6300$ and $8400$.