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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?

b. 4

c. 7

d. 5

e. 3

satisfy $\frac{1}{m} + \frac{4}{n}=\frac{1}{12}$,where n is an odd integer less than 60?

b. 4

c. 7

d. 5

e. 3

+3 votes

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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