218 is NOT divisible by 5 so the question, as stated, makes no sense.
No.
No.
All multiples of 218 (which are infinite) including 218, 436, 654, 872, 1090 . . .
The product of the prime factors of 218 is 218. The prime factors of 218 are 2 and 109, making their product 2 x 109 = 218! I wouldn't expect it to be otherwise: the product of the prime factors of a number equal the number.
Yes, any number is divisible by any other number
Any one of the infinite set of numbers of the form 3,237,441,264*k where k is an integer.
345 and 218
No. 436 is only divisible by: 1, 2, 4, 109, 218, 436.
As an even number greater than 2, 218 is divisible by 2 and therefore composite. Its positive integer factors are: 1, 2, 109, 218
NO. 1110 is not divisible by 9. A number is divisible by 9 if the sum of its digits is divisible by 9. 1110 = 1 + 1 + 1 + 0 = 3 (3 is not divisible by 9, thus, 1110 is not divisible by 9.)
81: 8 + 1 = 9 which is divisible by 9, so 81 is divisible by 9 162: 1 + 6 + 2 = 9 which is divisible by 9, so 162 is divisible by 9 199: 1 + 9 + 9 = 19 → 1 + 9 = 10 → 1 + 0 = 1 which is not divisible by 9, so 199 is not divisible by 9. 1125: 1 + 1 + 2 + 5 = 9 which is divisible by 9, so 1125 is divisible by 9. So 199 is the only one not divisible by 9.