# Bài Toán chia hết trong đề thi APMOPS 2013

Question 1: A 5-digit number written in the form $\overline{24abc}$ has the last three digits unknown. If this number is divisible by 3,4 and 5 respectively, find the greatest possible value that $\overline{abc}$ can take.

Solution:

This number ($A$) is divisible by 4 and 5. Thus $c$ must be 0. We can start with the case $a=9$. Because $A$ is divisible by 4, we imply that there are only four possible cases for $b$: 2,4,6,8. Now it is not difficult to conclude that the number 960 is our answer (notice that $A$ is divisible by 3).

Question 2: The sum of 10 positive integers, not necessary distinct, is 1001. If $d$ is the greatest common divisor of the 10 numbers, find the maximum possible value of $d$.

Solution:

Assume that these 10 numbers are $a_1,a_2,...,a_{10}$ and $a_i=dk_i$ $\forall i=1,...,10.$ We have $d(k_1+k_2+...+k_{10})=1001=7.11.13.$

Notice that $k_1+...+k_{10} \geq 10$ and it’s smallest possible value is 11 (it is possible because we can take $k_1=k_2=...=k_9=1$ and $k_{10}=2$). In this case we have $d=7.13=91.$ Hence, the maximum possible value of $d$ is 91.