$ N $ is a positive integer and $ p $ and $ q $ are primes. If $ N=p q $ and $ \frac{1}{N}+\frac{1}{p}=\frac{1}{q}, $ then find the value of $ N $.

\( N \) is a positive integer and \( p \) and \( q \) are primes. If \( N=p q \) and \( \frac{1}{N}+\frac{1}{p}=\frac{1}{q}, \) then find the value of \( N \).

Solution:

If \( N=p q \) and \( \frac{1}{N}+\frac{1}{p}=\frac{1}{q}, \) =>

 \( \frac{1}{p q}+\frac{1}{p}=\frac{1}{q}, \)

=>

\( \frac{1 + q}{p q} =\frac{1}{q}, \) =>

p = 1 + q

There are only two prime numbers that satisfy this condition that is 2 and 3; q = 2, p = 3

Therefore N = pq = 3 x 2 = 6

So N = 6 is the answer

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Updated on: 10-Oct-2022

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