2, 5 and 10 are factors of 120 because they divide evenly into 120 with no remainder.
120 is divisible by 1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 20, 24, 30, 40, 60 and 120.
No, 590 is divisible by 2, 5 and 10.
It is divisible by 2, 5 and 10 but not the rest.
Any element of the set of numbers of the form *k, where k is an integer, is evenly divisible. However, 308 is not evenly divisible by either.
no not by 10 as 216 ends in 6 not 0 same with 5 as does not end in 5 or 0 216 = 2 x 108 = 3 x 72 = 4 x 54 = 6 x 36 = 9 x 24 --------------------------------------- Last digit of 216 is even (one of {0, 2, 4, 6, 8}) so 216 is divisible by 2 6 + 1 + 2 = 9 which is divisible by 3 so 216 is divisible by 3 Last 2 digits are 16 which are divisible by 4, so 216 is divisible by 4 Last digit is not 0 nor 5, so 216 is NOT divisible by 5 216 is divisible by both 2 and 3 so 216 is divisible by 6 6 + 1 + 2 = 9, so 216 is divisible by 9 Last digit is not 0, so 216 is NOT divisible by 10 216 is divisible by 2, 3, 4, 6, 9. 216 is NOT divisible by 5 nor 10.
120 is divisible by the following numbers: 1 2 3 4 5 6 8 10 12 15 20 24 30 40 60 120.
The sum of its digits is 1+2+0 = 3 which is divisible by 3. So 120 is divisible by 3. The number 120 ends in a 0 and so it is divisible by 5.
Since 120 ends with zero, then 120 is divisible by 5.
It is divisible by 2 and 3. It isn't divisible by 5 and 10.
because 10 is also divisible by 2 and 5.
30, 60, 90, 120, 150, 180...
Because 10 is divisible by both 2 and 5
No, 120 is not the smallest number that is divisible by 1,2,3,4,5 and 6.
10 / 2 = 5 10 / 5 = 2
Yes, it is divisible by 2, 3 and 9, but is not exactly divisible by 5 or 10.
120 is divisible by 1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 20, 24, 30, 40, 60 and 120.
9 is divisible by 3 10 is divisible by 2, 5 and 10. Everything else, no.