CAT 2007 | Question: 21

Question

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

• A. $6$
• B. $4$
• C. $7$
• D. $5$
• E. $3$

Comments

this question came in IIITH pgee 2019.

Answer 1

$\frac{1}{m}$ = $\frac{1}{12}$ - $\frac{4}{n}$

$\frac{1}{m}$ = $\frac{n-48}{12n}$

$m = \frac{12n}{n-48} \\= \frac{2^2.3.n}{n-48}$ 

$n> 48$ and odd and $m$ is an integer.

thus n = 49, 51, 57 as for $n = 53,55,59$ we won't get integer values for $m$

Hence option (E), 3 is the Answer.

Answer 2

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