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Re: If a is divisible by 7!, then a/5 must be
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29 Nov 2017, 09:05

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Top Contributor

Bunuel 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

-----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)