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.