der of any number divided by 3. For example, 3,212
equals 3 ×1000 + 2 ×100 + 1 ×10 + 2 ×1 and so leaves
a remainder of 3 ×1 + 2 ×1 + 1 ×1 + 2 ×1 = 8 when
divided by 3. This is the sum of its digits. Of course, a
remainder of 8 is equivalent to a remainder of 2. We
have:
The remainder of any number divided by 3 is
the sum of its digits. Thus a number is divisi-
ble by 3 only if the sum of its digits is a multi-
ple of 3.
One can make repeated use of this rule to check for
divisibility. For example, 55,837 leaves a remainder of
5 + 5 + 8 + 3 + 7 = 28 when divided by 3. This corre-
sponds to a remainder of 2 + 8 = 10, which, in turn, is
a remainder of 1 + 0 = 1.
This rule shows that scrambling the digits of any
multiple of 3 produces a new number that is still a mul-
tiple of 3.
Divisibility by 4
As any multiple of 100 is divisible by 4, it suffices to
check whether or not the final two digits of a number
represent a two-digit multiple of 4. For example,
18,736 equals 187 ×100 + 36. As 36 is a multiple of 4
(it can be divided by 2 twice), we are sure then that
18,736 is a multiple of 4. We have:
A number is divisible by 4 if its final two digits
represent a two-digit number that can be
divided by 2 twice.
Divisibility by 5
Any number Ncan be written in the form N= 10a+ b
where bis the final digit in N. (For example, 739 = 73
×10 + 9.) As 10 is divisible by 5, we need only check
whether or not the final digit bis divisible by 5. This
gives:
A number is divisible by 5 only if its final digit
is 0 or 5.
Divisibility by 6
For a number to be divisible by 6 it must both be even
and a multiple of 3. This gives:
A number is divisible by 6 only if it is an even
number whose digits sum to a multiple of 3.
Divisibility by 7
Every number Nis of the form N= 10a+ b, that is, a
multiple of 10 plus a single digit. As 7a and 7b are
clearly divisible by 7, we have that 10a+ bleaves the
same remainder as 10a+ b– 7a– 7b= 3(a– 2b) does
when divided by 7. Thus it suffices to check whether
the quantity 3(a– 2b) is a multiple of 7. This can only
occur if a– 2bis a multiple of 7. Noting that ais the
original number Nwith its final digit removed and bis
the final digit of N, we have the rule:
To test whether or not a number is divisible by
7, remove the last digit and subtract twice that
digit from the number remaining. Then the
original number is divisible by 7 only if the
result of this operation is divisible by 7.
For example, to test whether or not 68,978 is divisible
by 7, remove the 8 and subtract twice this, 16, from
6,897, the number remaining. This gives 6,897 – 16 =
6,881. We can test whether or not 6,881 is a multiple
of 7 the same way: 6,881 →688 – 2 = 686, and once
more: 686 →68 – 12 = 56. That the final result, 56, is
divisible by 7 assures us that 68,978 is a multiple of 7.
Divisibility by 8
As any multiple of 1,000 is divisible by 8, it suffices to
check whether or not the final three digits of a number
represent a three-digit multiple of 8. For example,
648,728 equals 648 ×1000 + 728. As 728 can be
divided by 2 three times, and hence is a multiple of 8,
we have that 648,728 is divisible by 8.
A number is divisible by 8 if its final three dig-
its represent a three-digit number that can be
divided by 2 three times.
Divisibility by 9
Given that 10, 100, 1000,… all leave a remainder of
one when divided by 9, the divisibility rule for 9 is
identical to that of 3.
The remainder of any number divided by 9 is the
sum of its digits. Thus a number is divisible by 9
only if the sum of its digits is a multiple of 9.
Again, one may make repeated use of this rule. For
example, 76,937 leaves a remainder of 7 + 6 + 9 + 3 +
7 = 32 when divided by 9, which, in turn, corresponds
to a remainder of 3 + 2 = 5. This rule is often used to
divisibility rules 145