Question

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

Answer 1

I present here a trial solution.

Given

`1st` AP seies `1,8,15,22....`

And

`2nd` AP series `2,13,24,35 ......`

Let `n_1th` term of the first seies be a common intger with the `n_2 th` term of 2nd series.

So `1+(n_1-1)*7=2+(n_2-1)*11`

`=>7n_1+3=11n_2`

By trial we get

For `n_1=9,20,31,42....`

the corresponding values of `n_2=6,13,20,27.....`

we have `t_(n_1)=1+(n_1-1)*7`

So inserting values of `n_1` we get the following series of common terms for two series.

`1+(9-1)*7=57`

`1+(20-1).7=134`

`1+(31-1)*7=211`

....etc

obviously we get the same series by inserting the values of `n_2` in `t_(n_2)=2+(n_2-1)*11`

Hence common terms of both the series constitute an AP.
having first term `57` and common difference `77`

Let the last common integer of the series be the `nth` term of the series. This `nth` must be `<=2003`,the smaller last term of two given series.

Hence
`57+(n-1)*77<=2003`

`=>n<=2023/77`

`=>n<=26 21/77`

As `n` must be an integer.

`n=26`

Hence the number of common integers of two given serier is `n=26`