If a is divisible by 7!, then a/5 must be

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RULE: If integer N is divisible by integer d, we can write N = dk (for some integer k)
Example: 30 is divisible by 6, and we can write 30 = (6)(5).
Likewise, 100 is divisible by 4, and we can write 100 = (4)(25).
Or, if q is divisible by 5, and we can write q = 5k (where k is some integer).
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Given: a is divisible by 7!
So, a = (7!)(k) for some integer k
That is, a = (7)(6)(5)(4)(3)(2)(1)(k)

This means: a/5 = (7)(6)(5)(4)(3)(2)(1)(k)/5
Simplify to get: a/5 = (7)(6)(4)(3)(2)(1)(k)

Now let's examine the statements:
I. divisible by 6
a/5 = (7)(6)(4)(3)(2)(1)(k)
So, a/5 MUST be divisible by 6
Statement I is true

II. a multiple of 7
a/5 = (7)(6)(4)(3)(2)(1)(k)
So, a/5 MUST be divisible by 7
In other words, a/5 MUST be a multiple of 7
Statement II is true

III. a multiple of 10
a/5 = (7)(6)(4)(3)(2)(1)(k)
If k = 1, then a/5 is definitely NOT a multiple of 10, since we cannot see a 10 "hiding" in the prime factorization of a/5
Statement III need NOT be true

Answer: E

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Here's more information about the concept of divisors "hiding" in the prime factorization of a number:

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)

EVEN MORE HERE: