Generalised Form of Numbers Which Help to Deduce Divisibility Tests
If the digit at the tens place is a
and digit at the units place is b,
then the number is represented as 10a + b.
Similarly, a
three-digit number in generalized form is represented by 100a
+ 10b + c,
where a, b
and c are
the digits at the hundreds place, tens place and units place, respectively.
For example:
a. 875 = 8 × 100 + 7
× 10 + 5
b. abc =
a × 100 + b
× 10 + c,
i.e. 100a + 10b
+ c
Let you are
given a number 875 and you are asked to tell if it is divisible by 2, 3, 4, 5,
6, 8 or 10. By performing actual division, we find that it is divisible by 3
and 5 only. The process of performing actual division is time consuming. Let us
see if we can find some patterns which enable us to find whether a number is
divisible by 2, 3, 4, 5, 6, 8, 9, 10 and 11 by examining digits in the number
to be divided. Let us know about these patterns.
Divisibility Tests of Numbers
Divisibility Tests of Numbers by 2
The multiples of 2 are 2, 4, 6,
8, 10, 12 ... . Do you observe some pattern in the units place of these
numbers? In each number, the digit at the units place is even.
Thus, a
number is divisible by 2 if its units digit is 0, 2, 4, 6 or 8, i.e. an even
number.
Divisibility Tests of Numbers by 4
We know that 100 is completely
divisible by 4.
Take a number 732 and let us
find if it is completely divisible by 4.
Now, 732 = 7 × 100 + 32
Here, 700 is completely
divisible by 4 because 7 × 100 = 7 × 4 × 25.
Also, 32 is divisible by 4 as 32
÷ 4 = 8.
Now, any natural number of three
or more digits can be written in the form (a × 100 + b), where
a
is
a single-digit natural number and b is a
two-digit natural number.
A multiple of 100 is always
divisible by 4, the number formed by the last two digits of that number should
be divisible by 4.
Thus,
a
number with three or more digits is divisible by 4, if the number formed by its
digits at the tens place and units place is divisible by 4.
Divisibility of 1- or 2-digit numbers by 4 needs to
be checked by actual division.
Divisibility Tests of Numbers by 8
We know that 1000 is completely
divisible by 8.
Let us take a four-digit number 3464,
and let us see if it is completely divisible by 8.
Now, 3464 = 3 × 1000 + 464.
Here, 3 × 1000 is completely divisible
by 8 because 3 × 1000 = 3 × 8 × 125.
Further, 464 is divisible by 8
as 464 ÷ 8 = 58.
Hence, 3464 is completely
divisible by 8.
Any natural number with four or
more digits can be expressed in the form: (a × 1000 + b), where
a
is
a single-digit natural number and b is a
three-digit natural number.
Here, a
×
1000 is always divisible by 8. If b is
divisible by 8, the given number is divisible by 8.
Thus,
a
number with four or more digits is divisible by 8, if the number formed by the
last three digits of the number is divisible by 8.
The divisibility of 1-, 2- or 3-digit numbers by 8
needs to be checked by actual division.
Divisibility Tests of Numbers by 5
The multiples of 5 are 5, 10,
15, 20, 25, 30, ... . Do you find some common pattern?
Each of these numbers has 0 or 5
at the units place.
Think of some more numbers with
0 or 5 at the units place, say 235, 450, 645, 3100, etc. All these numbers are
divisible by 5.
Thus, a number
is divisible by 5 if its units digit is either 0 or 5.
Divisibility Tests of Numbers by 10
The multiples of 10 are 10, 20,
30, 40, ... . Do you find some common pattern?
Each of these numbers has zero
at the units place.
Think of some more numbers with
0 at the units place, say 400, 650, 8570, ... etc.
All the numbers are divisible by
10.
Thus, if a
number has 0 at the units place, then it is divisible by 10.
Divisibility Tests of Numbers by 3
Let us consider a four-digit
number with a
at
the thousands place, b at the hundreds place, c
at
the tens place and d at the units place. Thus, the
number (in expanded form) is expressed as:
1000a
+
100b
+
10c
+
d
= (999
+ 1)a
+
(99 + 1)b
+
(9 + 1)c
+
d
= (999x
+
99y
+
9z) + (a
+
b
+
c
+
d)
= 9(111a
+
11b
+
c) + (a
+
b
+
c
+
d)
9 is a factor of (999a
+
99b
+ 9c) ⇒ (999a + 99b
+ 9c) is
divisible by 3.
The given number will be
completely divisible by 3 only if a + b
+
c
+
d
is
divisible by 3.
Thus, a
number is divisible by 3, if the sum of its digits is divisible by 3.
Divisibility Tests of Numbers by 9
The multiples of 9 are 9, 18,
27, 36, 45, ... .
Let us see the pattern when the
digits of these numbers are added: 9, 1 + 8, 2 + 7, 3 + 6, 4 + 5, ... . All
these sums are divisible by 9.
Consider some other numbers
which are divisible by 9, say 84978, 333405 and 2734506. You observe a pattern
when you find the sum of all the digits:
8 + 4 + 9 + 7 + 8 = 36, 3 + 3 + 3
+ 4 + 0 + 5 = 18, 2 + 7 + 3 + 4 + 5 + 0 + 6 = 27
All these sums are divisible by
9.
Thus, a number
is divisible by 9, if the sum of its digits is divisible by 9.
Divisibility Tests of Numbers by 6
The multiples of 6 are 6, 12,
18, 24, 30, 36, 42, 48, 54, 60 ...
Do you observe that the digit at
the units place is always even and the sum of the digits is always divisible by
3.
Thus, a
number is divisible by 6, if the number is divisible by 2 as well as 3.
Divisibility Tests of Numbers by 11
Let us consider a four-digit
number with a
at
the thousands place, b at the hundreds place, c
at
the tens place and d at the units place.
Then the number in expanded form
is expressed as:
1000a
+
100b
+
10c
+
d
=
(1001 – 1)a
+
(99 + 1)b
+ (11
– 1)c
+
d
= (1001a
+
99b
+
11c) – a
+
b
–
c
+
d
= 11(91a
+
9b
+
c) + (b
+
d) – (a
+
c)
Here, the number will be
divisible by 11 if (b + d) – (a
+
c) is
either zero or a multiple of 11.
Thus, a
number is divisible by 11, if the difference between the sum of the digits at
odd places and the sum of the digits at even places is either zero or a
multiple of 11.
Divisibility Tests of Numbers by 7
Let us check if the number 4578 is
divisible by 7 or not.
Find the difference between
twice the digit at the units place and the number formed by the remaining
digits, i.e. 457 – 2 × 8 = 441, Again, 44 – 2(1) = 42, which is divisible by 7.
Thus, a number
is divisible by 7, if the difference between twice the units digit and the
number formed by remaining digits is either 0 or a multiple of 7.
Related Topics: