2004 remainder 2, 1002 remainder 1, 2005 remainder 3, multiplication remainder 6.

2.35 divided by 2, the first one is 37, and the second one is 72. 72 can be divisible by 3, so 72.

3.(42n+ 12)/2 1=2n remainder 12

4,4 * 5 * 6 = 1 20 divided by 7 and1

12 1 divided by 7 leaves 2.

We need to add five remainder 1, so it is121+120 * 5 = 721.

5. Description 125-90=35 can divide that number exactly.

That number may be 1, 5, 7, 35.

Try 35, it meets the requirements, so it is 35.

6. Find rules to find even numbers, odd numbers, odd numbers, even numbers, odd numbers, ... (This can actually be proved, but considering that the requirements for primary school students are not so high, I won't say it. )

100 must be an even number.

Look at the regularity of the remainder divided by 3: 0, 1, 0, 2, 0, 1, 0, 2, ... and then 100 remainder 2.

So the number100th is divided by 6 and 2.