# Look at the pattern of numbers below. In what row and column does the number 2008 sit?

## $\left(\begin{matrix}A & B & C & D \\ 8 & 6 & 4 & 2 \\ 10 & 12 & 14 & 16 \\ 24 & 22 & 20 & 18 \\ 26 & 28 & 30 & 32 \\ \vdots & \vdots & \vdots & \vdots\end{matrix}\right)$

column A (assumes 4 columns), row 251

#### Explanation:

We have a series of numbers that starts with:

$\left(\begin{matrix}A & B & C & D \\ 8 & 6 & 4 & 2 \\ 10 & 12 & 14 & 16 \\ 24 & 22 & 20 & 18 \\ 26 & 28 & 30 & 32 \\ \vdots & \vdots & \vdots & \vdots\end{matrix}\right)$

The pattern is that of a "snake" - with the initial number, 2, being in column D, then the pattern increases by 2 for each number moving from D to A, then drops down to the next row and increases by 2 again as it moves from A to D.

We're looking for the column that 2008 will appear in.

We can describe the numbers in the various rows and columns this way:

$\left(\begin{matrix}A & B & C & D \\ 2 + \left(2\right) 3 & 2 + \left(2\right) 2 & 2 + \left(2\right) 1 & 2 + \left(2\right) 0 \\ 2 + \left(2\right) 4 & 2 + \left(2\right) 5 & 2 + \left(2\right) 6 & 2 + \left(2\right) 7 \\ 2 + \left(2\right) 11 & 2 + \left(2\right) 10 & 2 + \left(2\right) 9 & 2 + \left(2\right) 8 \\ 2 + \left(2\right) 12 & 2 + \left(2\right) 13 & 2 + \left(2\right) 14 & 2 + \left(2\right) 15 \\ \vdots & \vdots & \vdots & \vdots\end{matrix}\right)$

And so with the pattern, we can find where 2008 will sit.

$2008 = 2 + 2 \left(n\right)$

$2 n = 2006$

$n = \frac{2006}{2} = 1003$

We can now divide by 8 - the remainder will tell us what column:

$\frac{1003}{8} = 125 \frac{3}{8} \implies \text{column A}$

With the column settled, let's now find the row. We can divide 2008 by 8:

$\frac{2008}{8} = 251$

We can see that 8 is on the first row, 16 on the second, 24 on the third, and so on. 2008 is evenly divisible 251 times, and so it's on the 251st row.

~~~~~
Another way to approach finding the column is to utilize bases, specifically base 8. Let's look at the chart we have, but in base 8 numbers:

$\left(\begin{matrix}A & B & C & D \\ 10 & 6 & 4 & 2 \\ 12 & 14 & 16 & 20 \\ 30 & 26 & 24 & 22 \\ 32 & 34 & 36 & 40 \\ \vdots & \vdots & \vdots & \vdots\end{matrix}\right)$

We can convert 2008 from base 10 to base 8 and use this chart to find our number

Changing from base 10, which has places such as $1 \left({10}^{0}\right) , 10 \left({10}^{1}\right) , 100 \left({10}^{2}\right)$, and so on, to base 8, which has #1 (8^0), 8 (8^1), 64 (8^3), and so on, involves finding the adjusted digits.

${8}^{3} = 512 \implies 512 \times \textcolor{red}{3} = 1536 \implies 2008 - 1536 = 472$
${8}^{2} = 64 \implies 64 \times \textcolor{g r e e n}{7} = 448 \implies 472 - 448 = 24$
${8}^{1} = 8 \implies 8 \times \textcolor{b r o w n}{3} = 24 \implies 24 - 24 = 0$

${2008}_{10} = \textcolor{red}{3} \textcolor{w h i t e}{00} \textcolor{g r e e n}{7} \textcolor{w h i t e}{00} \textcolor{b r o w n}{3} \textcolor{w h i t e}{00} 0$
$\textcolor{w h i t e}{00000000} \overline{{8}^{3}} \textcolor{w h i t e}{0} \overline{{8}^{2}} \textcolor{w h i t e}{0} \overline{{8}^{1}} \textcolor{w h i t e}{0} \overline{{8}^{0}}$

The last two digits are 30, and so that puts us in column A.

To find the row, we can take the digits, drop the final digit (because it's a 0, we're not in the middle of a series and not needing to add 1 additional row as a "final partial row", and re-evaluate the number:

$3 \left({8}^{2}\right) + 7 \left({8}^{1}\right) + 3 \left({8}^{0}\right) = 3 \left(64\right) + 7 \left(8\right) + 3 \left(1\right) = 251$

May 8, 2017

$2008$ is the first number in row $251$

#### Explanation:

We are given a pattern of numbers which are all just multiples of 2.

However, in the first row they run from right to left and in the second from left to right. This pattern then repeats.

$\left(\begin{matrix}A & B & C & D \\ 8 & 6 & 4 & 2 \\ 10 & 12 & 14 & \textcolor{b l u e}{16} \\ 24 & 22 & 20 & 18 \\ 26 & 28 & 30 & \textcolor{b l u e}{32} \\ \vdots & \vdots & \vdots & \vdots\end{matrix}\right)$

Every second line ends on a multiple of 16.

So find out how many groups of 16 can be obtained from $2008$

$2008 \div 16 = 125.5$

That means there are $125$ full groups of $\textcolor{b l u e}{16}$
The last row will end on $125 \times 16 = \textcolor{b l u e}{2000}$

But each group consists of $2$ rows.

$125 \times 2 = 250$ rows.

Row $250$ will therefore be:

$\left(1994 , 1996 , 1998 , \textcolor{b l u e}{2000}\right)$

There are four multiples of 2 needed to make $2008$

The four multiples of 2 in row $251$ will be:

$\left(\begin{matrix}A & B & C & D \\ 8 & 6 & 4 & 2 \\ 10 & 12 & 14 & 16 \\ 24 & 22 & 20 & 18 \\ 26 & 28 & 30 & 32 \\ \vdots & \vdots & \vdots & \vdots \\ 1994 & 1996 & 1998 & \textcolor{b l u e}{2000} \\ 2008 & 2006 & 2004 & 2002\end{matrix}\right)$

Jul 17, 2017

A different approach!

row 251
column 4

#### Explanation:

Consider the first row of 2 4 6 8 $\to$ 4 values.

So every full row will have 4 values

Notice that the last value of 8 divided 2 gives 4. Which is the count of values for row 1. Consequently the total count of values in the whole matrix is $2008 \div 2 = 1004$

$\textcolor{g r e e n}{\text{Count of values is 1004}}$

Now consider the first 3 rows of the content of the matrix

$\left(\begin{matrix}2 & 4 & 6 & 8 \\ 16 & 14 & 12 & 10 \\ 18 & 20 & 22 & 24\end{matrix}\right)$

Row 1: 1 is odd and counts left to right
Row 2: 2 is even and counts right to left

This will be repeated throughout all the rows
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{g r e e n}{\text{The total count of rows "=(1004" values")/("4 values per row")=251" rows}}$

$\textcolor{g r e e n}{251 \text{ is odd so it counts left to right}}$

$\textcolor{g r e e n}{\text{Thus 2008 is in the right most column:}}$

$\textcolor{g r e e n}{\text{row 251}}$
$\textcolor{g r e e n}{\text{column 4}}$