Question

An order pair (n,m) of positive integers satisfying 1n+1m−1mn=25 , the value of mn is
(20)
(25)
(40)
(15)​

Answer 1

Step-by-step explanation:

Multiplying both sides by  mn , we get  n+m−1=2mn5 . We can rearrange to get  n(1−2m5)=1−m , which, first multiplying on both sides by  5 , allows us to solve for  n  as follows:  n=5–5m5–2m=5m−52m−5 .

Rewrite this as  2n=10m−102m−5=10m−25+152m−5=5+152m−5 . What this tells us is that, since  m,n  must both be integers,  2m−5  must be a factor of  15 . Notice that as  15  is odd,  2m−5 , which can take on any odd value, can be any factor of  15 . Another constraint to watch out for is that the left hand side is  2n , which must be even, but since  15  is odd,  152m−5  will always be odd, which means  5+152m−5  is always even, so that constraint is taken out of the picture.

Now, we can iterate through all of the factors of  15 .

If  2m−5=15 ,  (m,n)=(10,3) .

If  2m−5=5 ,  (m,n)=(5,4) .

If  2m−5=3 ,  (m,n)=(4,5) .

If  2m−5=1 ,  (m,n)=(3,10) .

If  2m−5=−1 ,  (m,n)=(2,−5) . This is invalid as  n  must be positive according to the question.

If  2m−5=−3 ,  (m,n)=(1,0) . This is also invalid as  n  must be positive. In addition, even if that constraint didn’t exist, the original equation contained division by  n  so it clearly can’t be  0 .

If  2m−5=−5 ,  (m,n)=(0,1) . This is invalid as  m  must be positive according to the question, and, like the previous possibility,  m  can’t be  0  anyway.

If  2m−5=−15 ,  (m,n)=(−5,2) . This is also invalid as  m  must be positive.

Therefore, our pairs are  (10,3),(5,4),(4,5),(3,10) , making the answer  4 .

Edit: Corrected the values of  n  listed above (I forgot to take out the factor of  2  from  2n ).

Answer 2

Correct Question :- An order pair (n,m) of positive integers satisfying 1/n+1/m − 1/mn = 2/5 , the value of mn is

a) 20

b) 25

c) 40

d) 15

Solution :-

→ 1/n + 1/m - 1/mn = 2/5

multiplying both sides by 5mn

→ 5m + 5n - 5 = 2mn

→ 5m + 5n - 2mn = 5

→ 5m + n(5 - 2m) = 5

multiplying both sides by 2,

→ 10m + 2n(5 - 2m) = 10

adding and subtracting 25 in LHS,

→ 10m + 2n(5 - 2m) + 25 - 25 = 0

→ 10m - 25 + 2n(5 - 2m) = 10 - 25

→ 5(2m - 5) + 2n(5 - 2m) = (-15)

→ 5(2m - 5) - 2n(2m - 5) = (-15)

→ (2m - 5)(5 - 2n) = (-15)

→ (-1)(2m - 5)(2n - 5) = (-1) * 15

→ (2m - 5)(2n - 5) = 15

→ (2m - 5)(2n - 5) can be equal to (1 * 15) ,(15 * 1) , (3 * 5) and (5 * 3) .

therefore,

→ 2m - 5 = 1, 3 ,5 or 15 .

→ 2m = 6, 8, 10 or 20 .

→ m = 3, 4, 5 or 10 .

and,

→ 2n - 5 = 15, 5 ,3 or 1 .

→ 2n = 20, 10, 8 or 6 .

→ n = 10, 5, 4 or 3 .

hence,

→ Possible values of m * n = 10*3 = 30, 5*4 = 20, 4*5 = 20 or 3*10 = 30 .

Hence, From options (A) 20 is correct answer.