# Finding the nth term in a repeating number sequence

I'm trying to figure out how to solve these types of repeating number sequence problems. Here is one I made up:

Consider the following repeating number sequence: {4, 8, 15, 16, 23, 42, 4, 8, 15, 16, 23, 42, 4, 8, 15, 16, 23, 42,…} in which the first 6 numbers keep repeating. What is the 108th term of the sequence?

I was told that when a group of k numbers repeats itself, to find the *n*th number, divide n by k and take the remainder r. The *r*th term and the *n*th term are always the same. 108 / 6 = 18, r = 0 So the 108th term is equal to the 0th term? Undefined?

I'm confused at how this works.

Thanks!

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Think of your sequence as doubly infinite, $\dots,4,8,15,\dots$, and the zeroth term is the term just before the first term. PS: I just noticed what sequence you are using. Are you sure you're confused, not lost? –  Gerry Myerson Jul 22 '11 at 1:52
Yes, hahaha :) Thanks though, I understand now. –  stoicfury Jul 22 '11 at 1:57

You are looking for modular arithmetic. The procedure you described of dividing and taking the remainder is encapsulated in modular arithmetic.

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