SOLUTION: How many pairs of positive integers m,n satisfy 1/m + 4/n=1/12, where m is an odd integer less than 60?

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Question 1132291: How many pairs of positive integers m,n satisfy 1/m + 4/n=1/12, where m is an odd integer less than 60?
Found 2 solutions by MathLover1, greenestamps:
Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!
We are given that 1%2Fm+%2B+4%2Fn+=+1%2F12 such that m is odd and m%3C+60.
Let us try and find out the value of n in terms of m
=> 1%2Fm+=+1%2F12+-+4%2Fn
=> 4%2Fn+=+1%2F12+-+1%2Fm
=>4%2Fn+=+%28m+-+12%29%2F12m
=>4%2F%28%28m+-+12%29%2F12m%29=n
=>n=48m%2F%28m+-+12%29

So, we can say that m+%3E+12
Possible odd values of+m such that 12+%3C+m+%3C+60+ are

13+| 15 | 17 | 19 | 21 | 23 | 25+|27 | 29 | 31|33|35|37|39|41|43|45|47|49|51|53|55|57|+59| (24 integers)
check possible pairs:
m=+13, n+=++624
m+=+15, n+=+240
m+=+17,+n+=816%2F5 (irreducible)=> not an integer
same is if m=19=>irreducible
m+=+21, n+=+112
m+=+23,+n+=1104%2F11 =>irreducible
m+=+25, n+=1200%2F13 (irreducible)
m+=+27, n+=432%2F5 (irreducible)
m+=+29,+n+=1392%2F17 (irreducible)
m+=+31, n+=1488%2F19 (irreducible)
m+=+33, n+=528%2F7 (irreducible)
m+=+35, n+=1680%2F23 (irreducible)
and rest are irreducible

so, you have 3 pairs that satisfy given condition
m=+13,+n+=++624=>13,624
m+=+15,+n+=+240=>15,240
m+=+21, n+=+112=>21,112


Answer by greenestamps(13200) About Me  (Show Source):
You can put this solution on YOUR website!


Tutor @MathLover1 has the right answers.

Here is a way to find them without trying all the odd integers less than 60.

In her solution, she shows algebraically that

n+=+48m%2F%28m-12%29

Perform the division indicated by that expression:

n+=+48m%2F%28m-12%29+=+%2848m-576%2B576%29%2F%28m-12%29+=+48+%2B+576%2F%28m-12%29

In this form, 48 is an integer, and n has to be an integer; that means 576/(m-12) has to be an integer.

The problem requires m to be an odd integer; that means m-12 is an odd integer.

The prime factorization of 576 is (2^6)(3^2).

Therefore, for 576/(m-12) to be an integer, with m-12 odd, m-12 has to be a factor of 3^2.

So the only possible values of m-12 are 1, 3, and 9; that makes the possible values of m 13, 15, and 21.

m = 13 --> n = 48+576/1 = 48+576 = 624
m = 15 --> n = 48+576/3 = 48+192 = 240
m = 21 --> n = 48+576/9 = 48+64 = 112