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

1 Answer
Aug 1, 2018

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