### Legs Eleven

Take any four digit number. Move the first digit to the 'back of the queue' and move the rest along. Now add your two numbers. What properties do your answers always have?

### Why 8?

Choose any four consecutive even numbers. Multiply the two middle numbers together. Multiply the first and last numbers. Now subtract your second answer from the first. Try it with your own numbers. Why is the answer always 8?

##### Stage: 3 Challenge Level:

Ege, Burcu, Bana and Alara all sent in examples of numbers that they had experimented with. They found that their digits always added up to a multiple of nine, and that the numbers themselves were also divisible by nine. They concluded that there is a link between these two properties, so that if a number has digits that sum up to nine, it must be a multiple of nine. Here are some of their examples:

From the set of numbers from $1$ to $9$ and by using each number once and once only;

Example 1 : $345 + 6789 + 210 = 7344$ =816*9
$7 + 3 + 4 + 4 = 18$. $18$ is a multiple of $9$, so the sum is divisible by nine.

Example 2 : $1023 + 4 + 5 + 6 + 7 + 8 + 9 = 1062$ = 118*9
$1 + 0 + 6 + 2 = 9$. $9$ is a multiple of $9$, so the sum is divisible by nine.

Example 3: $1234 + 56 + 789 = 2079$ = 231*9
$2 + 0 + 7 + 9 = 18$. $18$ is a multiple of $9$, so the sum is divisible by nine.

Example 4: $6723 + 14589 = 21312$ = 2368*9
$2 + 1 + 3 + 1 + 2 = 9$. $9$ is a multiple of $9$, so the sum is divisible by nine.

They also repeated the exercise for the set of numbers $1-8$ and found that the result was the same:

Example 1 : $23 + 467 + 158 = 648$ = 72*9
$6 + 4 + 8 = 18$. 18 is a multiple of $9$, so the sum is divisible by nine.

Example 2: $123 + 45 + 67 + 8 = 243$ = 27*9
$2 + 4 + 3 = 9$. $9$ is a multiple of $9$, so the sum is divisible by nine.

Example 3: $6245 + 137 + 8 = 6390$ = 710*9
6 + 3 + 9 + 0 = 18. 18 is a multiple of 9, so the sum is divisible by nine.

Example 4: $154 + 786 + 32 = 972$ = 108*9
$9 + 7 + 2 = 18$. $18$ is a multiple of $9$, so the sum is divisible by nine.

And for the set of numbers $0-9$:

Example 1 : $1023 + 45 + 67 + 89 = 5679$ = 631*9
$5 + 6 + 7 + 9 = 27$. $27$ is a multiple of $9$, so the sum is divisible by nine.

Rohaan from Longbay Primary School explained why this always works for the sum of any numbers made from the digits $1-9$:

I think the reason behind this is when you add all the digits (from $1$ to $9$) the total is $45$. $45$ is divisible $9$ so whatever groups of numbers you make and add up must be divisible by $9$.

That's right, and the numbers $1-8$ add up to $36$, which is also a multiple of $9$, so the rule still works. For the sets of numbers $1-6$ and $1-5$, Ege and Banu found a similarly interesting result for multiples of $3$:

Example 1 : $231 + 4 + 65 = 300$ = 100*3
$3 + 0 + 0 = 3$. $3$ is a multiple of $3$, so the sum is divisible by three.

Example 2 : $12 + 34 + 56 = 102$ = 34*3
$1 + 0 + 2 = 3$. $3$ is a multiple of $3$, so the sum is divisible by three.

So there you have it! This rule only works for multiples of $3$ or $9$, but it makes it very quick and easy to find out whether or not a big number is divisible by $3$ or $9$ without using a calculator. Thank you for all your excellent solutions.