Solution:
We can write, 132 = 3 × 4 × 11
So, the number that is exactly divisible by 3, 4 and 11 is exactly divisible by 132
As we know, a number is divisible by 3, if the sum of its digits is divisible by 3
Also, a number is divisible by 4, if the number formed by its last two digits is divisible by 4
And, a number is divisible by 11, if the difference between the sum of its digits at odd places and the sum of its digits at even places is either 0 or 11
Considering number 26908,
⇒ Sum of digits = 2 + 6 + 9 + 0 + 8 = 25, which is not divisible by 3
∵ The number 26908 is not divisible by 3, it is not divisible by 132
Considering number 26912,
⇒ Sum of digits = 2 + 6 + 9 + 1 + 2 = 20, which is not divisible by 3
∵ The number 26912 is not divisible by 3, it is not divisible by 132
Considering number 26928,
⇒ Sum of digits = 2 + 6 + 9 + 2 + 8 = 27, which is divisible by 3
⇒ No. formed by last two digits = 28, which is divisible by 4
For divisibility by 11,
⇒ Difference = (2 + 9 + 8) – (6 + 2) = 19 – 8 = 11
∵ The number 26928 is divisible by 3, 4 & 11, it is divisible by 132
Considering number 26932,
⇒ Sum of digits = 2 + 6 + 9 + 3 + 2 = 22, which is not divisible by 3
∵ The number 26932 is not divisible by 3, it is not divisible by 132
© examsiri.com