Correct Answer - Option 3 : 266
Calculation:
Concept:
Divisibility law of 2 ⇒ A number divisible by 2 if its last digit is 0, 2, 4, 6, or 8.
Divisibility law of 3 ⇒ A number divisible by 3, if the sum of its digit is divisible by 3
Divisibility law of 5 ⇒ A number divisible by 5 if its last digit is 0 or 5
Given:
Number of digits divisible by 2 which starting from 2 to 1000 = Total 500 = A
Number of digits divisible by 3 which starting from 3 to 999 = Total 333 = B
Number of digits divisible by 5 which starting from 5 to 1000 = Total 200 = C
Number of digits divisible by 2 and 3 together which starting from 6 to 996 = Total 166 = AB
Number of digits divisible by 3 and 5 together which starting from 15 to 990 = Total 66 = BC
Number of digits divisible by 5 and 2 together which starting from 10 to 1000 = Total 100 = CA
Number of digits divisible by 2, 3 and 5 altogether which starting from 30 to 990 = Total 33 = ABC
So the count of numbers which are divisible by either 2, 3 or 5 from 1 to 1000 will be,
Now, the number of integers from 1 to 1000 that are not divisible by any of the numbers 2, 3, 5, is
= 1000 - |A| - |B| - |C| + |AB| + |AC| + |BC| - |ABC|.
= 1000 - (500 + 333 + 200 - 166 - 66 - 100 + 33)
= 1000 - (734)
= 266
Divisibility laws of 4 ⇒ A number is divisible by 4 if its last 2 digits divisible by 4.
Divisibility law of 6 ⇒ A number divisible by 6 if the number divisible by 2 and 3 both.
Divisibility law of 7 ⇒ Take the last digit of the number, double it Then subtract the result from the rest of the number If the resulting number is evenly divisible by 7, so is the original number divisible by 7.
Divisibility law of 8 ⇒ A number divisible by 8 if its last three-digit is divisible by 8
Divisibility law of 9 ⇒ A number is divisible by 9 if the sum of its digit is divisible by 9.
