Thursday, Jan 19th

Problem
How many 7-digit numbers are multiples of 388 and end in 388?

Solution
In order to a number $n$ be a multiple of 388 and to end in 388, we have to have that $n-388$ is also a multiple of 388, i.e. the problem reduces to find all 4 digit numbers $m$ such that $m10^3$ is a multiple of $388$.

Since $388=2^2\times 97$, $m$ has to be just a multiple of $97$. Thus $m=11\times97, 12\times 97\dots, 103\times 97$ which gives 93 possible numbers.