The numeral '21' is evenly divisible by 21, 7, 3, and 1.
They're both divisible by 1,3,7, and 21.
There is nothing that is divisible in 37 and 21. Because 37 is a prime number. However, 21 is divisible by 3 and 7.
Because 21 is a multiple of 1, 3, 7 and 21.
Because 21 is divisible by numbers other than 1 and 21. 21 is divisible by 1, 3, 7, 21
By any multiples of 21
a number is divisible by 12 if the number is also divisible by both 3 and 7. You mean a number is divisible by 21 if it is divisible by 3 and 7.
No, it is not.
No, it is not.
No, it is not.
They are members of the infinite set of numbers of the form 21*k where k is an integer.
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.