15 and 5 both go into every multiple of 15 .
510. To be divisible by 3, it must be a multiple of 3. Thus the required number is a multiple of both 3 and 5, which will be a multiple of their lcm: lcm(3, 5) = 15. → 495 ÷ 15 = 33 → first multiple of 15 greater than 495 is 15 x 34 = 510 → 525 ÷ 15 = 35 → last multiple of 5 less than 525 is 15 x 34 x 510 → number required is 510.
15, 30, 45
No.15 = 3 x 5So for 200 to be a multiple of 15, it must be a multiple of both 3 and 5:2 + 0 + 0 = 2 which is not a multiple of 3 so 200 is not a multiple of 3.200 ends in 0, so it is a multiple of 5.Thus 200 is a multiple of 5, but not a multiple of 3; so it is not a multiple of 15.
It is any one of the infinite set of numbers of the form 15*k where k is an integer.
15
15, obviously... reasoning: 3 times 5=15
number is both a prime number and a multiple of 5 = 5
It is not possible to give a sensible answer to this question. The least common multiple (LCM) refers to a multiple that is COMMON to two or more numbers. You have only one number in the question!
The number 5 is both a factor and a multiple of 5, because5 and 1 are both factors of 5.5, 10, 15, 20, 25... are multiples of 5. The common number is 5.
15 and 5 both go into every multiple of 15 .
Which number is a factor of 10 , but not a multiple of 5
510. To be divisible by 3, it must be a multiple of 3. Thus the required number is a multiple of both 3 and 5, which will be a multiple of their lcm: lcm(3, 5) = 15. → 495 ÷ 15 = 33 → first multiple of 15 greater than 495 is 15 x 34 = 510 → 525 ÷ 15 = 35 → last multiple of 5 less than 525 is 15 x 34 x 510 → number required is 510.
Numbers that are divisible by both 3 and 5 must be multiples of the least common multiple of 3 and 5, which is 15. Therefore, any number that is divisible by 15 will also be divisible by both 3 and 5. This includes numbers like 15, 30, 45, 60, and so on.
Any multiple of 15.
3 and 5 are factors of 3270
None of them are multiples of 15. The be a multiple of 15, a number must have both the prime factors 3 and 5. None of them have a prime factor of 5, thus none can be a multiple of 15.