Pick up a number, say 12. Write down a list of all its divisors, including the number itself.

In our case: 1,2,3,4,6,12. Now evaluate AA and HA- the arithmetic and the harmonic averages.

AA=(1+2+3+4+6+12)/6=28/6
HA=6/(1/1+1/2+1/3+1/4+1/6+1/12)=6*12/28

Now evaluate the product AA*HA: (28/6)*(6*12/28)

Surprise, surprise! - the answer is 12, a number picked by us.

Please explain why it works with any number.

The list of divisors, in this case, 1, 2, 3, 4, 6, 12, consists of 3 pairs of numbers that multiply to 12, 1*12, 2*6, and 3*4.

AA = (sum_divisors) / (num_divisors)
HA = (num_divisors) / (sum_inverse_divisors)

The sum of the inverses is the secret to it all. As every inverse will have a common denominator of our value, X, adding the terms simply reverses the order in which the factors show up. That is, 1/1 * 12/12 = 12/12, 1 * 12 = 12 ... 1/2 * 12 / 12 = 6 / 12, 2 * 6 = 12 ... 1/3 * 12/12 = 4 / 12, 3*4 = 12 ... etc.

So, the final sum of these inverses winds up being:

12/12 + 6/12 + 4/12 + 3/12 + 2/12 + 1/12 = 28 / 12 ... or (sum_divisors) / X

That leaves us with: HA = (num_divisors) / ((sum_divisors) / X)

Multiplying HA by AA gives the following:

((sum_divisors) / (num_divisors)) * ((num_divisors) / ((sum_divisors) / X))

As you can see, the sum_divisors cancel out, as do the num_divisors, leaving:

HA * AA = 1 / (1 / X), which reduces to X, the number with which we started.

I'm sorry about the weird formatting, but I didn't want to mess around with symbols and all that to show it correctly.

Edited on March 12, 2010, 3:40 pm

Posted by Justin on 2010-03-12 15:38:05