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To be divisible by 6, the number must be divisible by both 2 and 3:

To be divisible by 2 the last digit must be even, ie one of {0, 2, 4, 6, 8};

To be divisible by 3, sum the digits of the number and if this sum is divisible by 3, then the original number is divisible by 3.

As the test can be repeated on the sum, repeat the summing until a single digit remains; only if this number is one of {3, 6, 9} is the original number divisible by 3.

If the number is not divisible by 2 or 3 (or both) then the number is not divisible by 6.

examples:

  • 126
Last digit is even so it is divisible by 2

1 + 2 + 6 = 9 which is divisible by 3, so it is divisible by 3

→ 126 is divisible by both 2 and 3, so it is divisible by 6

  • 124
Last digit is even so it is divisible by 2

1 + 2 + 4 = 7 which is not divisible by 3, so it is not divisible by 3

→ 126 is divisible by 2 but not divisible by 3, so it is not divisible by 6

  • 123
Last digit is not even so it is not divisible by 2

We can stop at this point as regardless of whether it is divisible by 3 or not, it will not be divisible by 6. However, for completeness:

1 + 2 + 3 = 6 which is divisible by 3, so it is divisible by 3

→ 123 is divisible by 3 but not divisible by 2, so it is not divisible by 6

  • 121
Last digit is not even so it is not divisible by 2

We can stop at this point as regardless of whether it is divisible by 3 or not, it will not be divisible by 6. However, for completeness:

1 + 2 + 1 = 4 which is not divisible by 3, so it is not divisible by 3

→ 121 is not divisible by either 2 or 3, so it is not divisible by 6

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โˆ™ 2018-03-26 10:49:42
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Q: What are the divisibility rule for 6?
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