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

Aug 1, 2018

I present here a trial solution.

Given

$1 s t$ AP seies $1 , 8 , 15 , 22. \ldots$

And

$2 n d$ AP series $2 , 13 , 24 , 35 \ldots \ldots$

Let ${n}_{1} t h$ term of the first seies be a common intger with the ${n}_{2} t h$ term of 2nd series.

So $1 + \left({n}_{1} - 1\right) \cdot 7 = 2 + \left({n}_{2} - 1\right) \cdot 11$

$\implies 7 {n}_{1} + 3 = 11 {n}_{2}$

By trial we get

For ${n}_{1} = 9 , 20 , 31 , 42. \ldots$

the corresponding values of ${n}_{2} = 6 , 13 , 20 , 27. \ldots .$

we have ${t}_{{n}_{1}} = 1 + \left({n}_{1} - 1\right) \cdot 7$

So inserting values of ${n}_{1}$ we get the following series of common terms for two series.

$1 + \left(9 - 1\right) \cdot 7 = 57$

$1 + \left(20 - 1\right) .7 = 134$

$1 + \left(31 - 1\right) \cdot 7 = 211$

....etc

obviously we get the same series by inserting the values of ${n}_{2}$ in ${t}_{{n}_{2}} = 2 + \left({n}_{2} - 1\right) \cdot 11$

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 $n t h$ term of the series. This $n t h$ must be $\le 2003$,the smaller last term of two given series.

Hence
$57 + \left(n - 1\right) \cdot 77 \le 2003$

$\implies n \le \frac{2023}{77}$

$\implies n \le 26 \frac{21}{77}$

As $n$ must be an integer.

$n = 26$

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