VJesus12 wrote:If a is divisible by 7!, then a/5 must be
I. divisible by 6
II. a multiple of 7
III. a multiple of 10
A. I, II, and III
B. None
C. II only
D. III only
E. I and II only
-----ASIDE---------------------
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).
-----ONTO THE QUESTION!---------------------
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
Cheers,
Brent
-----ASIDE---------------------
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)
