Page 47 - Revised Maths Wisdom Class - 6

Prime Time 45

ILLUSTRATIONS

Example 1: Check the divisibility of 1030801 by
(a) 3 (b) 5 (c) 6 (d) 9 (e) 10 (f) 11
Solution: (a) Sum of all the digits = 1 + 0 + 3 + 0 + 8 + 0 + 1 = 13
13 is not divisible by 3

Thus, 1030801 is not divisible by 3.
(b) Since the ones place digit is 1 (not 0 or 5)
Thus, 1030801 is not divisible by 5

(c) As 1030801 is not divisible by 3 and 2, it is not divisible by 6.
(d) Sum of all the digits = 1 + 0 + 3 + 0 + 8 + 0 + 1 = 13 which is not divisible by 9.
Thus, 1030801 is not divisible by 9
(e) As ones place digit is not 0, it is not divisible by 10.

(f) 1 0 3 0 8 0 1

Sum of odd place digits = 1 + 3 + 8 + 1 = 13
Sum of even place digits = 0 + 0 + 0 = 0

13 – 0 = 13 which is not divisible by 11.
Thus, 1030801 is not divisible by 11.

Example 2: In each of the following replace ‘*’ by the smallest possible digit to make them divisible by the
given number.
(a) 478* by 2 (b) 4*59948 by 11

Solution: (a) As we know that a number is divisible by (b) 4 * 5 9 9 4 8
2 if its last digit is 0, 2, 4, 6 or 8. 4 + 5 + 9 + 8 = 26
So the smallest digit that can replace * is 0. * + 9 + 4 = 13 + *

Hence, 4780 is the required number.
Their difference = 26 – (13 + *)

= 26 – (13 + 2)
Example 3: Is 16415 divisible by 7? = 26 – 15 = 11
Solution: Given number is 16415 Hence 2 is the required digit.

1641 – (5 × 2) = 1641 – 10 = 1631 This is a hit and trial method and we try
163 – (1 × 2) = 163 – 2 = 161 with smallest whole number onwards i.e.,
16 – (1 × 2) = 16 – 2 =14 0, 1, 2, ... etc.
Thus, 14 is divisible by 7
Hence, 16415 is divisible by 7.

Example 4: A number is divisible by 5 and 13 both. By which other number will that be always divisible?
Solution: 5 and 13 are co-primes as they have no common factor other than 1.
According to the rule if a number is divisible by two co-prime numbers, then it is also divisible by
their product. Hence it is also divisible by 5 × 13 = 65