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24 Sep 2021, 04:45
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E
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
45%
(medium)
Question Stats:
67% (01:49) correct
33%
(02:00)
wrong
based on 307
sessions
History
Date
Time
Result
Not Attempted Yet
If the least common multiple of a positive integer n and 50 is 100, what is the sum of all possible values n ?
A. 4
B. 20
C. 104
D. 120
E. 124
M37-02
A. 4
B. 20
C. 104
D. 120
E. 124
M37-02
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18 Nov 2021, 02:32
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Official Solution:
If the least common multiple of a positive integer \(n\) and 50 is 100, what is the sum of all possible values \(n\) ?
A. \(4\)
B. \(20\)
C. \(104\)
D. \(120\)
E. \(124\)
Given: \(LCM(n, \ 50)=100\);
\(LCM(n, \ 2*5^2)=2^2*5^2\).
\(n\) must be a multiple of \(2^2\) (because 50 is not and the LCM is). Also, the only other prime \(n\) can have is 5 (because 100 does not have any other prime but 2).
So, \(n\) can be \(2^2=4\), \(2^2*5=20\) and \(2^2*5^2=100\).
The sum of all possible values \(n\) is therefore \(4+20+100=124\).
Answer: E
If the least common multiple of a positive integer \(n\) and 50 is 100, what is the sum of all possible values \(n\) ?
A. \(4\)
B. \(20\)
C. \(104\)
D. \(120\)
E. \(124\)
Given: \(LCM(n, \ 50)=100\);
\(LCM(n, \ 2*5^2)=2^2*5^2\).
\(n\) must be a multiple of \(2^2\) (because 50 is not and the LCM is). Also, the only other prime \(n\) can have is 5 (because 100 does not have any other prime but 2).
So, \(n\) can be \(2^2=4\), \(2^2*5=20\) and \(2^2*5^2=100\).
The sum of all possible values \(n\) is therefore \(4+20+100=124\).
Answer: E
General Discussion
24 Sep 2021, 06:30
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Bunuel
n can equal 4, 20 or 100
Sum = 4 + 20 + 100 = 124
Answer: E
