Page 67 - ICSE Math 8

Solution: Number (abc) = 694
On reversing the digits, number (cba) = 496
2
abc – cba = 99(a – c) = 3 × 11 × (a – c)
Now, a = 6, b = 9, c = 4
2
2
\ 694 – 496 = 3 × 11 × (6 – 4) = 3 × 11 × 2
(a) We know that abc – cba when divided by 33 gives quotient as 3(a – c).
\ 694 – 496 when divided by 33, will give quotient as 3 × (6 – 4) = 3 × 2 = 6.
(b) abc – cba when divided by 99, gives quotient as a – c.
\ 694 – 496 when divided by 99, will give quotient as 6 – 4 = 2.
(c) abc – cba when divided by a – c, gives quotient as 99.
\ 694 – 496 when divided by 2, will give quotient as 99.

EXERCISE 5.1

1. Without actual calculations, find the quotient when 58 + 85 is divided by: (a) 13 (b) 11
2. Without actual calculations, find the quotient when 73 – 37 is divided by: (a) 9 (b) 4
3. If the sum of the number 842 and two other numbers obtained by arranging the digits of 842 in cyclic
order is divided by 111, 14 and 37 respectively, what will be the quotient in each case?

Tests of Divisibility

Test of Divisibility by 2
A number is divisible by 2 if its units digit is 0, 2, 4, 6 or 8.
For a number in the generalized form:
(a) A 2-digit number 10a + b is divisible by 2 if b is 0, 2, 4, 6 or 8.
(b) A 3-digit number 100a + 10b + c is divisible by 2 if c is 0, 2, 4, 6 or 8.
For example, 64, 82, 170, 256, 638, etc., are divisible by 2.
Test of Divisibility by 3

A number is divisible by 3 if the sum of its digits is divisible by 3.
For a number in the generalized form:
(a) A 2-digit number 10a + b is divisible by 3 if a + b is divisible by 3.
(b) A 3-digit number 100a + 10b + c is divisible by 3 if a + b + c is divisible by 3.
For example, 27, 39, 126, 234, 729, etc., are divisible by 3.
Test of Divisibility by 4
A number is divisible by 4 if the number formed by its ones and tens digits is divisible by 4 or if its ones and
tens digits are both ‘zeros’.
For a number in the generalized form:
A 3-digit number 100a + 10b + c is divisible by 4 if bc is divisible by 4 or b and c are both 0.
For example, 32, 56, 300, 524, 936, etc., are divisible by 4.

Test of Divisibility by 5
A number is divisible by 5 if its ones digit is either 0 or 5.
For a number in the generalized form:
(a) A 2-digit number 10a + b is divisible by 5 if b is either 0 or 5.
(b) A 3-digit number 100a + 10b + c is divisible by 5 if c is either 0 or 5.
For example, 15, 65, 90, 135, 280, etc., are divisible by 5.

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