GeminiHeat wrote:
If M is the least common multiple of 90,196, and 300, which of the following is NOT a factor of M?
A. 600
B. 700
C. 900
D. 2,100
E. 4,900
-----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 a multiple of k, then k is "hiding" within the prime factorization of NConsider these examples:
24 is a multiple of
3 because 24 = (2)(2)(2)
(3)Likewise, 70 is a multiple of
5 because 70 = (2)
(5)(7)
And 112 is a multiple of
8 because 112 = (2)
(2)(2)(2)(7)
And 630 is a multiple of
15 because 630 = (2)(3)
(3)(5)(7)
-----ONTO THE QUESTION!------------------------------------------------------------------------------------------------
Given: M = the least common multiple of 90, 196, and 300This means
M is a multiple of 90, 196, and 30090 = (2)(3)(3)(5)
This means the prime factorization of M must include one 2, two 3's and one 5.
So far, we know that:
M = (2)(3)(3)(5)(?)(?)(?) [There may be additional values in the prime factorization of M]196 = (2)(2)(7)(7)
This means the prime factorization of M must include two 2's and two 7's
Since the prime factorization of M already includes one 2, we need to add one more 2.
Since a prime factorization of M doesn't include any 7's yet, we must add two 7's to get:
M = (2)(2)(3)(3)(5)(7)(7)(?)(?)(?) 300 = (2)(2)(3)(5)(5)
This means the prime factorization of M must include two 2's, one 3 and two 5's
The prime factorization of M already includes two 2's, so we have that covered.
The prime factorization of M already includes one 3, so we have that covered.
Since the prime factorization of M already includes one 5, we need to add one more 5 to get:
M = (2)(2)(3)(3)(5)(5)(7)(7) At this point, we can see that m is now a multiple of 90, 196, and 30
So,
M = (2)(2)(3)(3)(5)(5)(7)(7)Now check the answer choices...
A) 600600 = (2)(2)(2)(3)(5)(5)
For 600 to be a factor of M, there must be three 2's, one 3 and two 5's "hiding" in the prime factorization of M. Since, M only has two 2's in its prime factorization, 600 is NOT a factor of M.
Answer: A