1, 3, 9, 13, 39, 117.
117
939 is evenly divisible by 3 939 is divisible by 8, but not evenly, quotient is 117, remainder is 3
None of these numbers are prime: 91 is divisible by 7 117 is divisible by 3 84 and 156 are both divisible by 2
117 is a composite number and not a prime number because it is divisible by 1, 3, 9, 13, 39, 117
117 is not a prime number. It is divisible by 3. 3*39=117.
The answer respectively is 117. 117 divided by 3 equals 39 and 117 is not divisable by 2. 105, 111, 117.... keep adding 6 so you always have an odd number, and still divisible by 3
117 / 3 = 39 39 / 3 = 13 Giving us prime factors of 3, 3 and 13. Meaning it's divisible by: 3 3 * 3 = 9 3 * 13 = 39 13 So 117 is divisible by 3, 9, 13, and 39
Yes it is.
No, 117 is only divisible by: 1, 3, 9, 13, 39, 117.
3x117=513 or 513/3=117 or 513/117=3
1, 3, 9, 13, 39, 117.
Neither number is prime: 117 is divisible by 3 and 84 is divisible by 2.
117
yes, 351/3=117
939 is evenly divisible by 3 939 is divisible by 8, but not evenly, quotient is 117, remainder is 3
Yes it is - 39 times