Add the digits of the number together and if the sum is divisible by 3 then the original number is divisible by 3.
The test can be applied to the sum and so the summation can be repeated until a single digit remains; if this digit is 3, 6 or 9 then the original number is divisible by 3.
the answer 3,6,9,24,and 27.
Divisibility by 48 requires divisibility by 3 AND by 48Divisibility by 3 requires tat the digital root (sum of digits) is divisible by 3.Divisibility by 16 requires that the number formed by the last four digits is divisible by 16.Of both these are satisfied by a number then it is divisible by 48.
there is a divisibility for 24 the rule is you can divide 24 as 6 and 4 i think
it is divisible by 3 and 9
3 and 9. And they divide into 123456789 whether or not you use divisibility rules!
No.
It is divisibility by 3 and divisibility by 5.Divisibility by 3: the digital root of an integer is obtained by adding together all the digits in the integer, with the process repeated if required. If the final result is 3, 6 or 9, then the integer is divisible by 3.Divisibility by 5: the integer ends in 0 or 5.
It is: 5034/3 = 1678
It is 3 6 9
Divisibility by 48 requires divisibility by 3 AND by 48Divisibility by 3 requires tat the digital root (sum of digits) is divisible by 3.Divisibility by 16 requires that the number formed by the last four digits is divisible by 16.Of both these are satisfied by a number then it is divisible by 48.
It is divisible by 3 (3*125 = 375).
678 is divisible by 3.
By using the divisibility rule for 3, we find that 117 is divisible by 3. That makes it composite.
The divisibility rule for 3 is add up the digits and check for divisibility. 7+2+4 = 13.13 is not evenly divisible by 3 so 724 is not divisibleby 3.
there is a divisibility for 24 the rule is you can divide 24 as 6 and 4 i think
it is divisible by 3 and 9
Not equally. The decimal answer is 30.66666
3 - 31