50th Mathematical Olympiad in Ukraine, Third Round (January 23, 2010)

Answer: a) yes; b) no.

It's obvious that a number received by Andrew is divisible by $16$ and by $125$, because from $16$ consecutive numbers more than four are divisible by $2$ and at least $3$ are divisible by $5$. It also implies that three last digits in Andrew's number are $0$.

a) Let the numbers $a,a+1,a+2,\dots ,a+15$ be written on the board. Then Olesya obtained the number $8(2a+15)$. Putting $a=55$ we have that three last digits of this number are $0$, so the answer to the case a) is 'yes'.

b) Since the number obtained by Andrew is divisible by $16$, then it's true for the number obtained by four last digits of this number. If we assume that the answer to b) is 'yes' then the number obtained by four last digits of the number $8(2a+15)$ is divisible by $16$. It follows that $8(2a+15)÷16$. This contradiction gives the answer 'no'.