One approach is to test each answer choice (starting with the smallest answer choice) to find the smallest value that's divisible by 2, 3, 4, 5, 6, 8, and 9

Another option is to BUILD the number in question. Let's build the number...

-----ASIDE---------------------
A lot of integer property questions can be solved using prime factorization.
For questions involving divisibility, divisors, factors and multiples, we can say:
If N is divisible by k, then k is "hiding" within the prime factorization of N

Consider these examples:
24 is divisible by 3 because 24 = (2)(2)(2)(3)
Likewise, 70 is divisible by 5 because 70 = (2)(5)(7)
And 112 is divisible by 8 because 112 = (2)(2)(2)(2)(7)
And 630 is divisible by 15 because 630 = (2)(3)(3)(5)(7)
-----ONTO THE QUESTION!---------------------

Let's say that N = the smallest integer that is divisible by 2, 3, 4, 5, 6, 8, and 9
So if N is divisible by 2, we know that a 2 must be hiding in its prime factorization. That is N = (2)(?)(?)(?)(?)(?)
Likewise, if N is divisible by 3, there must be a 3 hiding in the prime factorization of N. That is N = (2)(3)(?)(?)(?)(?)
Likewise, if N is divisible by 4, there must be two 2's hiding in the prime factorization of N. We already have one 2, so we need to add one more 2 to get: N = (2)(3)(2)(?)(?)(?)
Since N is divisible by 5, we need to add a 5 to get: N = (2)(3)(2)(5)(?)(?)
In order to be divisible by 6, there must be one 2 and one 3 in the prime factorization of N. Since we already have one 2 and one 3, we can be certain that N is already divisible by 6
To be divisible by 8, there must be three 2's hiding in the prime factorization of N. We already have two 2's, so we need to add one more 2 to get: N = (2)(3)(2)(5)(2)(?)(?)
Finally, to be divisible by 9, there must be two 3's hiding in the prime factorization of N. We already have one 3, so we need to add one more 3 to get: N = (2)(3)(2)(5)(2)(3)

So, N = (2)(3)(2)(5)(2)(3) = 360

Answer: C

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