Which of the following numbers is divisible by 2, 5, 10 and 11?
Show answer & explanation
Correct answer: C
Explanation
The correct answer is option 3 i.e. 589270.
Divisible by 2: A number is divisible by 2 if its last digit is even. The last digit of 589270 is 0, which is even, so 589270 is divisible by 2.
Divisible by 5: A number is divisible by 5 if its last digit is 0 or 5. The last digit of 589270 is 0, so 589270 is divisible by 5.
Divisible by 10: A number is divisible by 10 if its last digit is 0. Since the last digit of 589270 is 0, it is divisible by 10.
Divisible by 11: A number is divisible by 11 if the difference between the sum of the digits in odd positions and the sum of the digits in even positions is a multiple of 11 (including 0).
Odd positions (1st, 3rd, 5th, 7th): 5, 9, 7, 0 → sum = 5 + 9 + 7 + 0 = 21
Even positions (2nd, 4th, 6th): 8, 2, 0 → sum = 8 + 2 + 0 = 10
Difference: 21 - 10 = 11, which is divisible by 11.
The answer is 589270 is divisible by all of these numbers.