Question

Find the number of ordered pairs of natural numbers (a,b) such that 1/a+1/b =1/2013.​

Answer 1

Answer:

Hope it helps you!

Have a good day

mark me as brainlliest!

Answer 2

13.

Given:

The equation is  as follows:

1/a+1/b =1/2013.​

To find:

The number of ordered pairs of natural numbers (a,b) such that 1/a+1/b =1/2013.​

Let's write the equation as follows:

ab=2013b+2013a

ab − 2013a − 2013b + $2013^{2}$ = $2013^{2}$

(a−2013)(b−2013)=$2013^{2}$

Hence, here a−2013  and  b−2013 must be complementary factors $2013^{2}$.

Since the $2013^{2}$=$3^{2}$⋅$11^{2}$⋅$61^{2}$ has  2⋅3⋅3⋅3 = 54 factors.

Then a>b. We can't use  (a−2013,b−2013)=(2013,2013)  & (−2013,−2013).

The remaining,  54−2=52  factors form the 52/2 = 26 pairs of complementary factors.

For each of the pair,  a−2013  must be the larger factor and  y−2013  must be the smaller factor. This gives 26  solutions  (a,b)  with  a>b .

Note, this allows solutions where  b  is negative. If we restrict a,b  to be positive integers [natural numbers], then there are only  13  solutions.

Hence the answer is 13.

#SPJ3