They are all the multiples of 15 including itself
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Since both 3 and 5 are prime numbers, only numbers that are multiples of its product are the numbers that are divisible by both. 15 is the LCM of 3 and 5 and hence all multiples of 15 are divisible by both 3 and 5
All whole numbers are divisible by 1. Numbers are divisible by 2 if they end in 2, 4, 6, 8 or 0. Numbers are divisible by 3 if the sum of their digits is divisible by 3. Numbers are divisible by 4 if the last two digits of the number are divisible by 4. Numbers are divisible by 5 if the last digit of the number is either 5 or 0. Numbers are divisible by 6 if they are divisible by 2 and 3. Numbers are divisible by 9 if the sum of their digits is equal to 9 or a multiple of 9. Numbers are divisible by 10 if the last digit of the number is 0.
yes it is divisible by all of those numbers
It is divisible by all of those numbers.
-- All of the numbers that end with a zero are divisible by 10.-- All of those, and all of the numbers that end with a 5, are also divisible by 5.