can 156 be divisible by 2,3,5,9 and 10?

The term "divisible" means that the quotient is an integer and the remainder is 0.

156 ÷ 2 = 78
156 ÷ 3 = 52
156 ÷ 5 = 31 ...... remainder = 1
156 ÷ 9 = 17 ...... remainder = 3
156 ÷ 10 = 15 ...... remainder = 6

Hence, 156 is divisible by 2 and 3,
but not divisible by 5, 9 and 10.

Can 156 be divisible by 2,3,5,9 and 10?


Well obviously 2 because the end digit ends in 6...and is an even number.

For 3 indeed it is because the sum of digits (1 + 5 + 6 = 12) is a multiple of 3...

For 5 no it is not because it doesn't end in 0 or 5...

For 9 it is not...no it is not since the sum of digits (12) isn't divisible by 9...

For 10 heck no, because the last digit isn't 0...

just by 2&3

No, 156 divisible by 2,3 only

Just by 2&3

It is only divisible by 2 and 3

yes, but only 2 and 3 evenly.

In order for a certain number to be a factor of a certain number, it must go into the number evenly.
(Don't mistake the hyphens as negative signs.)

156
------
1-156
2-78
3-52
4-39
6-26
12-13

2 yes as it is even number
10 no as it does not end in0
5 no as it does not end in 0 or 5
3 yes 15/3 =5 6/ = 2
9 no 15/9 1 remainder 6 66 does not divide by 9

shouldnt take a genius to figure out that only numbers ending in '0' are divisible by 10

If it's a question where decimals are acceptable then yes.