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21 is not divisible by 2 nor 5, but it is divisible by 3.
To be divisible by 2 the last digit must be even (one of {0, 2, 4, 6, 8}; the last digit 1 is not one of these, thus 21 is not divisible by 2
To be divisible by 3 sum the digits and if this total is divisible by 3, then so is the original number; the test can be repeated on the sum, so by repeatedly summing the digits of the totals until a single digit remains only if that single digit is 3, 6 or 9 is the original number divisible by 3. 21 → 2 + 1 = 3 which is divisible by 3 so 21 is divisible by 3.
To be divisible by 5, the last digit must be 0 or 5. The last digit of 21 is 1 which is not 0 nor 5, thus 21 is not divisible by 5.
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What is a harshad in math?
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