Adding all nine
Make a set of numbers that use all the digits from $1$ to $9$, once and once only.
For instance, we could choose:
$638, 92, 571$ and $4$
Add them up:
$638 + 92 + 571 + 4 = 1305$
$1305$ is divisible by $9$ (it is $145\times 9$)
(use a calculator to check this if you do not know yet how to divide by $9$)
Add each of the digits in the number $1305$ . What is their sum?
Or, perhaps we could choose:
$921, 4357$ and $68$
Add them up:
$921 + 4357 + 68 = 5346$
$5346$ is divisible by $9$ (it is $594 \times 9$)
Now try some other possibilities for yourself!
I wonder what happens if we use all $10$ digits from $0$ to $9$, once and once only?
Try some for yourself!
What do you think would happen if we used the eight digits from $1$ to $8$?
Test your hypothesis by trying some possibilities for yourself!
Were you correct?
Is there a pattern beginning to emerge? Do you have theory that might explain what is happening?
Try some different sets of digits for yourself!
See how many $2$-digit numbers you can find that are divisible by $9$.
What happens if you just use the numbers from $1$ to $8$?
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 2: $123 + 45 + 67 + 8 = 243$ = 27*9
Example 3: $6245 + 137 + 8 = 6390$ = 710*9
Example 4: $154 + 786 + 32 = 972$ = 108*9
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.