Question

If the average of a number N and its reciprocal is 1, what is the number N?

Answer 1

the reciprocal of N is obviously \(\frac 1N\)

so the average is 1 you say?

\[\large \frac{N + \frac{1}{N}}{2} = 1\]
\[\large N + \frac 1N = 2\]
\[\large \frac{N^2+ 1}{N} = 2\]
\[\large N^2 + 1 = 2N\]
\[\large N^2 - 2N + 1 = 0\]
solve for N

Answer 2

so n=1

Answer 3

yup