# The number of common integers for two arithmetic progressions 1,8,15,22,...2003 and 2,13,24,35,...2004 is?

##### 1 Answer

I present here a trial solution.

Given

And

Let

So

By trial we get

For

the corresponding values of

we have

So inserting values of

....etc

obviously we get the same series by inserting the values of

Hence common terms of both the series constitute an AP.

having first term

Let the last common integer of the series be the

Hence

As

Hence the number of common integers of two given serier is