Is there a number that is divisible by the numbers 1 to 9?

Question

I know that the number being divided would have to be even and that it would also have to therefore end in a zero to fascillitate the five.

If such a number exists what is it called?
It would, virtually, be the complete opposite of a prime number.

Answer 1

Yes, there are infinite such numbers that are divisible by the numbers 1 to 9.
Please note that 9*8*7*5 = 2520 is the smallest such number, since 9*8*7*5 is multiple of all numbers from 1 to 9.
Hence, any integer n = 2520k, with k be any positive integer, can meet your requirement.
This number does not have any particular name.
A more challenge problem would be:
Find a 9-digits number, using 1,2,3,4,5,6,7,8, and 9 as its digits only once and at least once, such that the m-digit number truncated from the last m-digits of this 9-digit number is divisible by m. For example, 312 is such 3-digit number, which uses 1, 2, and 3 as its digits. the last digit 2 is divisible by 1, the last two digits 12 is divisible by 2, and the last 3-digit number 312 is divisible by 3.

Answer 2

It is called 9! or nine factorial
1X2X3X4X5X6X7X8X9=362,880
It would therefore be divisible by 10 and each number below ten

No even number except 2 can be a prime number.
No number which ends in 5, can be a prime number
(except 5 itself)
If the digits in a number add up to nine or a multiple of nine, the number will be divisible by nine and cannot be a prime number.