# With a series of numbers 1, 2, 4, 7, 11, 16,... what is the remainder of the 2008th number in the series when divided by 5?

Remainder = 4

#### Explanation:

We have a series of numbers: 1, 2, 4, 7, 11, 16,...

and we're looking for the remainder of the 2008th number in the series is divided by 5.

We're dealing with a recursive addition series where:

${x}_{n + 1} = {x}_{n} + n$

And the first way I can think of to tackle this is to set up a spreadsheet and calculate it that way.

And so:

${x}_{1} = 1$
${x}_{2} = 1 + 1 = 2$
${x}_{3} = 2 + 2 = 4$
${x}_{4} = 4 + 3 = 7$
$\vdots \textcolor{w h i t e}{00} \vdots \textcolor{w h i t e}{00} \vdots \textcolor{w h i t e}{00} \vdots$

${x}_{2008} = 2 , 013 , 022 + 2007 = 2 , 015 , 029$

$\frac{2 , 015 , 029}{5} = 403 , 005.8 \implies \text{remainder} = 0.8 \times 5 = 4$

~~~~~

Another way to approach this is to look solely at the 1s digit and what happens when divided by 5.

For instance, our first number is 1. When divided by 5, the remainder is 1. The second number is 2. When divided by 5, the remainder is 2. And so on. Perhaps there is a pattern we can spot in our numbers:

$\left(\begin{matrix}\text{Number in series" & x & "/5" & "remainder} \\ 1 & 1 & .2 & 1 \\ 2 & 2 & .4 & 2 \\ 3 & 4 & .8 & 4 \\ 4 & 7 & 1.4 & 2 \\ 5 & 11 & 2.2 & 1 \\ 6 & 16 & 3.2 & 1 \\ 7 & 22 & 4.4 & 2 \\ 8 & 29 & 5.8 & 4 \\ \vdots & \vdots & \vdots & \vdots\end{matrix}\right)$

And a pattern sets up in the remainder: 1, 2, 4, 2, 1 and then repeat (I verified this in my spreadsheet - it holds!)

Since this repeats every 5 elements, we can look simply at the last digit of the order number (8 of the 2008) and see that the remainder will be 4.