Which of the following numbers is divisible by 2, 5, 10 and 11?

Question

Accepted Answer

To check divisibility by 2, 5, 10, and 11, the number must satisfy all these conditions:

1. Divisible by 2: The last digit must be even (0, 2, 4, 6, 8).
2. Divisible by 5: The last digit must be 0 or 5.
3. Divisible by 10: The last digit must be 0.
4. Divisible by 11: The difference between the sum of digits in odd positions and the sum of digits in even positions must be 0 or divisible by 11.

Check each number:

1. 203467
- Last digit is 7 (not 0), so not divisible by 2, 5, or 10.

2. 830942
- Last digit is 2 (even), so divisible by 2.
- Last digit not 0 or 5, so not divisible by 5 or 10.

3. 589270
- Last digit is 0, so divisible by 2, 5, and 10.
- Check divisibility by 11:
  Positions (from right to left):
  Digits: 5 8 9 2 7 0
  Odd positions: 0 (pos 1), 9 (pos 3), 5 (pos 5) sum = 0 + 9 + 5 = 14
  Even positions: 7 (pos 2), 2 (pos 4), 8 (pos 6) sum = 7 + 2 + 8 = 17
  Difference = 17 - 14 = 3 (not divisible by 11)

Re-check positions from left to right (standard method):
  Digits: 5(1),8(2),9(3),2(4),7(5),0(6)
  Sum odd positions (1,3,5): 5 + 9 + 7 = 21
  Sum even positions (2,4,6): 8 + 2 + 0 = 10
  Difference = 21 - 10 = 11, which is divisible by 11.

So 589270 is divisible by 11.

4. 1234560
- Last digit 0, so divisible by 2, 5, and 10.
- Check divisibility by 11:
  Odd positions: 1 + 3 + 5 + 0 = 9
  Even positions: 2 + 4 + 6 = 12
  Difference = 12 - 9 = 3 (not divisible by 11)

Therefore, only 589270 is divisible by 2, 5, 10, and 11.

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