Must I Do All This?!

We have $a_1 = 1$ and we want $a_n = 2015$ for some value of n.

We see that the difference b/w the first and the last term is 2014 and we know that $a_n = a_1 +(n-1)d$ . So if we have to now represent 2014 as a product of two numbers which shall be our $n-1$ and $d$ .

Like 1 and 2014. If you put d=1, you get n-1 = 2014 and vice versa.

Factoring 2014 we get $2014= 2*19*53$ .

So we have

$2014=1*2014$

$2014=2*1007$

$2014= 19*106$

$2014=53*38$

So we now know that d can attain values $1,2,19,38,53,106,1007,2014$

Summing them up we get 3240. Thus our answer is $\boxed {240}$

If N satisfies the condition then N needs to be a factor of 2015-1=2014

$2014= 2 \times 19 \times 53$

So values of N that satisfy the condition are

• 1
• 2
• 19
• 38
• 53
• 106
• 1007
• 2014

There sum is 3240

$3240= 3 \times 1000 +240$

So the answer is $\boxed{240}$

If a number N satisfies the condition in the problem, N must divide 2015-1=2014. The sum of all the divisors of 2014 is 3240. 3240 mod 1000 is 240.