All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars
 perplexus dot info

 In Harmony with averages (Posted on 2010-03-12)
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.

 No Solution Yet Submitted by Ady TZIDON No Rating

Comments: ( Back to comment list | You must be logged in to post comments.)
 A solution that might make sense. | Comment 1 of 2
The divisors of n come in pairs whose product is n:
call them 1 & n, d1&d2, d3&d4, etc...
[if n is a perfect square there will be and unpaired divisor sqrt(n)]
lets say there are a divisors total.

AA = (1+d1+d3+...+[sqrt(n)]+...+d4+d2+n)/a

HA = a/(1/1+1/d1+1/d3+...+[1/sqrt(n)]+...+1/d4+1/d2+1/n)
these fractions be given a common denominator and added:
=a/((n+d2+d3+...+[sqrt(n)]+...+d3+d1+1)/n)
=an/(n+d2+d3+...+[sqrt(n)]+...+d3+d1+1)
=an/(1+d1+d3+...+[sqrt(n)]+...+d4+d2+n)

AA*HH= (1+d1+d3+...+[sqrt(n)]+...+d4+d2+n)/a*an/(1+d1+d3+...+[sqrt(n)]+...+d4+d2+n)
=n

 Posted by Jer on 2010-03-12 15:07:40

 Search: Search body:
Forums (0)