25. Appending three digits at the end of 2007 , one obtains an integer $N$ of seven digits. In order to get $N$ to be the minimal number which is divisible by 3,5 and 7 simultaneously, what are the three digits that one would append?
25. Ans: 075 .
It suffices to make it divisible by 105 after appending. As $2007000=105 \times 19114+30$, the least number that we need to add is 75 .