If where ​, what is the value of n?

1+6+11+16+21+26+\cdots+91+96+101 \equiv n\pmod{15},

 

n=6

If 1 + 6 + 11 + ... + 96 + 101  =  n (mod 15), what is the value of n?

 

I believe that Guest's answer of 6 is correct; this is my slow, step-by-step way of finding that answer.

 

I plan to find the sum of the terms, divide that answer by 15, and the remainder will be n.

 

The series is an arithmetic series.

 

The formula for the sum of an arithmetic series is:  Sum  =  n(f + l) / 2

where  n = number of terms,  f = first term,  and  l = last term.

f = 1  and  l = 101

but I still need to find the number of terms.

 

The formula for the n^th term of an arithmetic sequence is:  t_n =  t₁  + (n - 1)d

t₁ = first term = f = 1

t_n = last term = l = 101

d = common difference = 5

Substituting these into the formula:  t_n =  t₁  + (n - 1)d     --->     101  =  1 + (n - 1)·5

                                                                    Solving:                   100  =  (n - 1)·5

                                                                                                      20  =  n - 1

                                                                                                         n = 21  (there are 21 terms in the series)

Now, solving for the sum:  Sum  =  n(f + l)/2

                                          Sum  =  21(1 + 101) / 2

                                          Sum  =  1071

 

Since we're working with mod 15, let's divide the sum by 15; the remainder will be our answer.

     1071 / 15  =  71, and the remainder is 6.

 

Therefore, our answer is 6.