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20 Apr 2019, 01:56
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D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
(hard)
Question Stats:
46% (02:26) correct
54%
(02:40)
wrong
based on 130
sessions
History
Date
Time
Result
Not Attempted Yet
How many pairs of positive integers m,n satisfy \(\frac{1}{m}+\frac{4}{n}=\frac{1}{12}\), where n is an odd integer less than 60
A. 5
B. 11
C. 6
D. 3
E. 4
A. 5
B. 11
C. 6
D. 3
E. 4
20 Apr 2019, 10:42
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nick1816
\(\frac{1}{m}+\frac{4}{n}=\frac{1}{12}\) can be rewritten as : \(m = \frac{12n}{n-48}\)
if m is a positive integer, then n must be >48.
since n is odd < 60, so the possible values of n are (49,51,53,55,57,59)
now test, which values of n maintain the condition that m is an integer:
if n = 49, \(m =\frac{12*49}{1}\) --> integer
if n = 51, \(m =\frac{12*51}{3}\) --> integer
if n = 53, \(m =\frac{12*53}{5}\) --> not integer
if n = 55, \(m =\frac{12*55}{7}\) --> not integer
if n = 57, \(m =\frac{12*57}{9}\) --> integer
if n = 59, \(m =\frac{12*59}{11}\) --> not integer
so only three pairs of n and m can satisfy the given conditions : D
General Discussion
26 May 2019, 18:22
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nick1816
Silly mistake, at the end didn't check whether n is I or not, good sum.
I did it differently but the trick is to find m is integer or not saves a lot of time.
