The harmonic mean of two positive integers is the reciprocal of the arithmetic mean of their reciprocals. For how many ordered pairs (x, y) of positive integers is the harmonic mean of x and y equal to 20? 🍺

threepointonefourone Jul 11, 2024

#1**0 **

Just realized the answer. I got confused on whether (35, 14) and (14, 35) should be counted as 1 or 2

threepointonefourone Jul 11, 2024

#2**+1 **

We can set variables to solve this problem.

First, let's let the reciprocals be \( 1/x ,1/y \)

The arithmetic mean of the two numbers are \( [ 1/x + 1/y] / 2 = [ x + y] / [ 2xy]\)

This means the harmonic mean is \( [ 2xy] / [ x + y] \)

Now, we write an equation. Since the harmonic mean must eqaual 20, we set the equation up.

Now, we put x in terms of y. We have

\(2xy / [ x + y ] = 20 \\ xy / [ x + y] = 10 \\ xy = 10x + 10y \\ yx - 10y = 10x y ( x - 10) = 10x \\ y = 10x / [ x - 10]\)

I counted 7 numbers. Not sure if that's right.

Thanks! :)

NotThatSmart Jul 11, 2024

#3**+2 **

I counted

(20, 20)

(35, 14)

(30, 15)

(60, 12)

(110, 11)

(11, 110)

(12, 60)

(15, 30)

(14, 35)

threepointonefourone
Jul 11, 2024

#6**0 **

I graphed it, and made sure for all integers x from 11 - 20 (10 doesn't work) that there is no integer B such that 1/10 - 1/x = 1/B

threepointonefourone Jul 12, 2024