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GRE Prep Club Team Member
Joined: 20 Feb 2017
Posts: 2506
Given Kudos: 1053
GPA: 3.39
If M is the least common multiple of 90,196, and 300, which of the fol
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18 Apr 2022, 09:59
18 Apr 2022, 09:59
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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
A. 600
B. 700
C. 900
D. 2,100
E. 4,900
Retired Moderator
Joined: 10 Apr 2015
Posts: 6218
Given Kudos: 136
If M is the least common multiple of 90,196, and 300, which of the fol
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18 Apr 2022, 13:26
18 Apr 2022, 13:26
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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
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 N
Consider 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 300
This means M is a multiple of 90, 196, and 300
90 = (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) 600
600 = (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
Re: If M is the least common multiple of 90,196, and 300, which of the fol
[#permalink]
09 May 2022, 03:18
09 May 2022, 03:18
GreenlightTestPrep wrote:
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
A. 600
B. 700
C. 900
D. 2,100
E. 4,900
The least common multiple of 90, 196, and 300 = (2)(2)(3)(3)(5)(5)(7)(7)
So, M = (2)(2)(3)(3)(5)(5)(7)(7)
Now check the answer choices...
A) 600
600 = (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
Well... i didn't even understand the first step even
A detailed explanation would be appreciated.
