Page 44 - SM inner class 8.cdr

Test of Divisibility place and c is in unit’s place, it can be written
Divisibility by 2:
Take the multiples of 2 : i.e. 2, 4, 6, 8, 10, 12,
14, 16, 18, 20, ………….etc
In these numbers the unit’s digit ends with
0,2,4,6, 8.
If the unit’s digit of the number is 0 or 2 or 4 or
6 or 8 ( even number ) then it is divisible by 2.
Otherwise it will not be divisible by 2.
Let us see the logic behind this rule.
If we take a three digit number 100 × a + 10 × b
+ c
where a is in hundred’s place, b is in ten’s place
and c is in unit’s place , then it can be written
as 100a + 10b + c = 2(50a + 5b) + c, 2(50a + 5b)
is multiple of 2. If the given number is divisible (a +
by 2, it is possible only if the unit’s
digit c = 0 or 2 or 4 or 6 or 8 ( even number).
Divisibility by 3 :
If we take a three digit number 100a+10b+c
where ‘a’ is in hundred’s place, ‘b’ is in ten’s
place and ‘c’ is in unit’s place.
100a + 10b + c = (99 + 1)a + (9 + 1) + c = 99a +
9b +(a + b + c)
= 9(11a + b) + (a + b + c) ® sum of given digits
9(11a + b) multiple of 3 .The given number is
divisible by 3 , only if the sum of the digits
(a + b + c) is multiple of 3 or (a+b+c) is divisibly by 3. 10(10a + b) + c
Then only, we can say that 100a + 10b + c is
divisible by 3.
Divisibility by 5 :
Take the multiples of 5. These are 5,10,15,
20,25,30,35 ,40,45,50,………….etc
In these numbers the unit’s digit is ‘0’ or ‘5’
If the units digit of a number is ‘0’ or ‘5’ then it
is divisible by 5.
Let us see the logic behind this rule .
If we take a three digit number 100a + 10b + c
where ‘a’ is in hundred’s place, b is in ten’s
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Mathematics In Focus - 8
as 100a + 10b + c = 5(20a + 2b) + c
5(20a + 2b) is multiple of 5.
The given number is divisible by 5, only if the
unit’s digit c = 0 or 5
Divisibility by 9
If we take a three digit number 100a+10b+c
where ‘a’ is in hundred’s place, ‘b’ is in ten’s
place and ‘c’ is in unit’s place.
100a + 10b + c = (99 + 1)a + (9 + 1) + c = 99a + 9b
+(a + b + c) = 9(11a + b) + (a + b + c)
® sum of given digits
9(11a + b) multiple of 9 .The given number is
divisible by 9, only if the sum of the digits
b + c) is multiple of 9 or (a + b + c) is
divisible by 9.
Then only, we can say that 100a + 10b + c is
divisible by 9.
Divisibility by 10:
Take the multiples of 10: 10, 20, 30, 40, 50, 60,
………….etc
In all these numbers the unit’s digit is ‘0’
Therefore if the unit digit of a number is
‘0’,then it is divisible by 10.
If we take a three digit number where ‘a’ is in
hundred’s place, ‘b’ is in ten’s place and ‘c’ is in
unit’s place can be written as 100a + 10b + c =
10(10a + b) is multiple of 10. If ‘c’ is a multiple
of 10 then the given number will be divisible
by 10. It is possible only if c = 0.
Example 11. Is 723 divisible by 3?
Solution: We could use the “2” rule: 7 + 2 + 3 =
12, and 12 ÷ 3 = 4 exactly divisible. Yes
Example 12. Is 314159 divisible by 9?
Solution: 314159 has a digit sum of 3 + 1 + 4 +
1 + 5 + 9 which is 23. 2 + 3 has a digit sum of 5,
And, since this single digit is not 9,
Then 314159 is not a multiple of 9. No