Page 40 - SM inner class 8.cdr

+ 4 % 1 ¼ 3 ÷ ¾

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Chapter5 6 ½ –

Playing with Numbers

When you start playing with numbers, they [Grouping a’ and b’s together]
(
give you new fun and new knowledge every =11a + 11b =11 a + b)
time. They have exciting patterns with which As we can see that 11 is a multiple of ab + ba,
so it is always divisible by 11.
they form puzzles and tricks. Let us have fun
and knowledge with numbers. Similarly, if we take the difference of ab – ba
a
-
= (10a + ) (10b + )
b
Now we shall learn to study numbers, writing
=10a + b -10b - a
2 and 3 digit numbers in general form, games
= (10a - ) (10b - )
-
a
b
and puzzles and deduction of divisibility test of
= 9a - 9b = 9 a b)
-
(
large numbers in general form.
As we can see that 9 is a multiple of ab - ba, so
Generalized Form of Numbers it is always divisible by 9.
Consider the number 564
3-digit number
564 = 500 + 60 + 4
= 5 100 + 6 10 + 4 General form of a 3-digit number is 100 × a +
´
´
10 × b + 1 × c.
This is the expanded form of a 3-digit
number. Now, shall learn to write 2-digit, 1. The difference between a 3-digit number
3-digit number in their general form using and a number obtained by reversing its
literal numbers a b, and c. digits is always divisible by 99.
Assume abc is a 3-digit number, where:
2-digit number
v a is the hundreds digit
The number in the general form can be written
as for example 26 = 2 ´ 10 + 6. General Form of v b is the tens digit
a 2-digit Number is 10 ´ a + 1 ´ b v c is the ones digit
1. The sum of a 2-digit number and the Generalised form of abc =100 × a + 10 × b + 1 × c
number obtained by interchanging its Example 1. Write the following numbers in
digits is always divisible by 11. generalised form.
2. The difference between a 2-digit number (i) 25 (ii) 73
and the number obtained by interchan-
(iii) 129 (iv) 302
ging its digits is always divisible by 9.
Solution: (i) 25 = 10 × 2 + 5 (ab = 10 ×a + b)
Assume ab is a 2-digit number.
(ii) 73 = 10 × 7 + 3

v a is the tens digit (iii) 129 = 100 × 1 + 10 × 2 + 9
v b is the ones digit (abc = 100 × a + 10 × b + c)
ab = 10 × a + 1 × b. (iv) 302 = 100 × 3 + 10 × 0 + 2

(Now, reversing the number it becomes ‘ba’) Example 2. Write the following in the usual
form.
ba = 10 ´ b + 1 ´ a
Now, ab + ba = (10a + 1b) + (10b + 1a) (i) 10 × 5 + 6 (ii) 100 × 7 + 10 × 1 + 8
= (10a + a) + (10b + b) (iii) 100 × a + 10 × c + b

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Mathematics In Focus - 8