No.
To be divisible by 6, the number must be divisible by both 2 and 3.
To be divisible by 2 the number must be even, ie its last digit must be one of {0, 2, 4, 6, 8}; 105's last digit is 5 which is odd so 105 is not divisible by 2.
To be divisible by 3, sum the digits of the number and if the result is divisible by 3, then so is the original number. For 1-5: 1 + 0 + 5 = 6 which is divisible by 3 therefore 105 is divisible by 3.
Although 105 is divisible by 3 it is not divisible by 2, thus it is not divisible by 6.
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