A number theory problem by Maxwell Johnson

The $n$ th term of this sequence is such that $a_{n} = a_{n-1} + 2^{n-2}$ for $n \ge 2$ with $a_{1} = 3$ . So the general formula for $a_{n}$ , with $n \ge 2$ , is

$a_{n} = 3 + \sum_{k=0}^{n-2} 2^{k} = 3 + (2^{n-1} - 1) = 2^{n-1} + 2$ .

(In fact this formula holds true for $n = 1$ as well.)

So $a^{13} = 2^{12} + 2$ . Now since $2^{3} \equiv 1 \mod{7}$ we have that, modulus $7$ ,

$(2^{12} + 2) \equiv ((2^{3})^{4} + 2) \equiv (1 + 2) \equiv 3 \mod{7}$ .

So the desired remainder is $\boxed{3}$ .

The difference between two terms goes on being multiplied by 2. So this way the 13 term is 4098. When you divide it by 7 remainder is 3

Note the pattern of the remainders in this sequence. 3, 4, 6, 3, 4, 6, .... Thirteenth term would have a remainder of 3.