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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 $60?$

1.  4
2.  7
3.  5
4.  3
edited | 204 views

1/m+4/n=1/12

(n+4m)/mn=1/12

mn=12(n+4m)=12n+48m   ----1

RHS will be even and it is given that n is odd so m should be even(even=even*odd).

n={1,3,5,...,57,59}

m={2,4,6,...,58,60}

by equation 1 we can conclude .. to get positive value of m n>48

so for  n=49

49m=12*49+48m

m=588 this is also even

n=51

51m-48m=12*51 => 3m=12*51 => m=204

n=53

5m=12*53=>m=127.2 which is not even integer

n=55

7m=12*55=> m=92.28 which is not even integer

n=57

9m=12*57=> m=76 this is valid

n=59

11m=12*59=>m=64.3

so number of possible solutions =3

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