Page 42 - Maths Class 06
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9 If the sum of the digits is divisible by 9. 1638 is divisible by 9 as the sum 1 + 6 + 3 + 8
= 18, which is divisible by 9.
10 If the last digit is 0. 7000 is divisible by 10 as last digit is 0.
11 If the difference of the sum of digits at even 172678 is divisible by 11 as (7 + 6 + 8) – (1 +
places and sum of digit at odd places is 0 or 2 + 7) = 21 – 10 = 11 which is divisible by 11.
divisible by 11.
12 If the number is divisible by both 3 and 4. 8316 is divisible by 12 as it is divisible by 3 as
well as 4.
EX AM PLE 1. Test whether the fol low ing num ber are di vis i ble by 3 or 9:
(a) 3271 (b) 8253
SO LU TION : (a) The given num ber is 3271.
The sum of digits = 3 + 2 + 7 + 1 = 13.
Since, 13 is not divisible by 3, so 3271 is not divisible by 3.
Since, 13 is not divisible by 9, so 3271 is not divisible by 9.
(b) The given number is 8253.
The sum of digits = 8 + 2 + 5 + 3 = 18.
Since, 18 is divisible by 3 and 9, therefore 8253 is divisible by both 3 and 9.
EX AM PLE 2. Is 69055 di vis i ble by 7?
SO LU TION : Con sider, 6905 - 5 2´ = 6905 10- = 6895
This rule can be again repeated to check divisibility.
Consider, 689 - 5 2´ = 689 - 10 = 679, which is divisible by 7.
Thus, 69055 is divisible by 7.
EX AM PLE 3. Which of the fol low ing num bers are di vis i ble by 5 and 10?
(a) 3527 (b) 13520 (c) 37595
SOLUTION : (a) In 3527, the digit at ones place is nei ther 0 nor 5. There fore, 3527 is not di vis i ble by
5 and 10.
(b) In 13520, the digit at ones place is 0. Therefore, 13520 is divisible by 5 and 10.
(c) In 37595, the digit at ones place is 5. Therefore, 37595 is divisible by 5. But the digit at
ones place is not 0, so 37595 is not divisible by 10.
EX AM PLE 4. Find the small est and great est digit ‘p’ such that 5p 729 is di vis i ble by 3.
SO LU TION : Consider the sum of dig its = 5 + p + 7 + 2 + 9 = 23 + p
24 is closest number to 23 that is divisible by 3
\ 23 + p = 24
Þ p = 1
\ 51729 is the required smallest number divisible by 3, satisfying the given condition.
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