Page 66 - ICSE Math 8

Example 2: Without actual calculations, find the quotient when 75 – 57 is divided by (i) 2 and (ii) 9.
Solution: Since 75 and 57 are numbers that can be obtained by reversing the digits of the other, so

-
-
Ê ab ba ab ba ˆ
ab
(75 – 57) ÷ 2 = 9 and (75 – 57) ÷ 9 = 2 Á  ab = 9 and 9 =- ˜
¯
Ë
-
Reversing the Digits—3-Digit Number
Let there be a 3-digit number abc. By changing the order of the digits we get the numbers bca and cab.
These numbers in the generalized form can be written as:
abc = (100 × a) + (10 × b) + c bca = (100 × b) + (10 × c) + a cab = (100 × c) + (10 × a) + b
On adding we get,
abc + bca + cab = (100a + 10b + c) + (100b + 10c + a) + (100c + 10a + b) = 111a + 111b + 111c
fi abc + bca + cab = 111(a + b + c) = 3 × 37 × (a + b + c)
So, it can be deduced that the sum of abc, bca and cab is exactly divisible by 3, 37, (a + b + c), 111,
3(a + b + c) and 37(a + b + c).
Number Divisor Quotient
abc + bca + cab 111 a + b + c
a + b + c 111
37 3(a + b + c)
3 37(a + b + c)
3(a + b + c) 37
37(a + b + c) 3
Now let’s try reversing the digits of a 3-digit number. Let there be a number abc. On reversing the order of
the digits, we get cba. In the generalized form
abc = 100a + 10b + c and cba = 100c + 10b + a
If a > c, then abc > cba, so
abc – cba = (100a + 10b + c) – (100c + 10b + a) = 99a – 99c = 99(a – c)
If c > a, then cba > abc, so
cba – abc = (100c + 10b + a) – (100a + 10b + c) = 99c – 99a = 99(c – a)
So, it can be deduced that the difference between abc and cba is always a multiple of 99 and the quotient is
equal to the difference between the hundreds and ones digits of the number.
Number Divisor Quotient
abc – cba 99 a – c
a – c 99
33(a – c) 3
3(a – c) 33
33 3(a – c)
3 33(a – c)
9 11(a – c)
11(a – c) 9
11 9(a – c)
9(a – c) 11

Example 3: Write the quotient when the difference between 694 and the number obtained by interchanging
its ones and hundreds digits is divided by:
(a) 33 (b) 99 (c) 2

54