+42 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? 🍺
+42 Just realized the answer. I got confused on whether (35, 14) and (14, 35) should be counted as 1 or 2
+1926 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! :)
+42 I counted
(20, 20)
(35, 14)
(30, 15)
(60, 12)
(110, 11)
(11, 110)
(12, 60)
(15, 30)
(14, 35)
+42 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
