Yes. 117/9=13.
An easy test for divisibility by nine (for an integer of any length) is to add all of the digits. If the sum is nine or a number divisible by nine, then the integer is divisible by nine. In this case, 1+1+7=9, so 117 is divisible by nine. (Be careful, this test fo divisibility only works generally for divisibility by three -- i.e., an integer is divisible by three if and only if the sum of its digits equals three or a multiiple of three-- and for nine.) To find if an integer is divisible by 4, you can check if the last two digits are. If they are, it is. To check if an integer is divisible by 6, you must make sure that the integer is divisible by both 3 and 2. If you want to check if an integer is divisible by 2, just make sure it's even. Any integer divisible by 10 will end in zero. Any integer divisible by 5 will end in either 5 or 0. This is an important part of pre-algebra, as well as algebra.
3, 9, 13, and 39.
117 and 225, yes. The rest, no.
No. 5 is not a factor of 117
According to my calculations, the following are evenly divisible by 9: 108, 117, 126, 135, 144, 153, 162, 171, 180, 189, and 198.
To determine what numbers 585 is divisible by, we need to find its factors. 585 is divisible by 1, 3, 5, 9, 13, 15, 39, 45, 65, 117, 195, and 585 itself. This means that 585 can be divided evenly by any of these numbers without leaving a remainder.
Yes. 117 is evenly divisible by nine.
117 divided by 9 = 13
1, 3, 9, 13, 39, 117.
Yes. 117 is evenly divisible by nine.
No, 117 is only divisible by: 1, 3, 9, 13, 39, 117.
117
3, 9, 13, and 39.
No it doesn't because 9 times 13 is 117.
117 and 225, yes. The rest, no.
117 is a composite number and not a prime number because it is divisible by 1, 3, 9, 13, 39, 117
109 is divisble by 9
117