Let n be some arbitrarily huge number; call each prime between 1 and n a 'GIRL'; call each even semiprime between 1 and n a 'BOY'.

Prove that there are two GIRLs for every BOY.

(An even semiprime is a composite number one of whose two prime factors is 2.)

Since the solution seems rather counterintuitive, as demonstrated by some of the earlier comments, I append the ratios of BOYs to GIRLs up to 10^n:

BOY  GIRL  Ratio
2  4  0.5
15  25  0.6
95  168  0.56547619
669  1229  0.544344996
5133  9592  0.535133445
41538  78498  0.529159979
348513  664579  0.524411695
3001134  5761455  0.520898627
26355867 50847534 0.518331272
234954223 455052511 0.516323319
2119654578 4118054813 0.514722284
19308136142 37607912018 0.513406225
etc.

Which leads naturally to the question, how huge is 'arbitrarily huge' - answer, pretty enormous:

(log(n))/(2 log(n/2)) =
0.501:   n~~6.54678x10^150
0.5001:  n~~2.82493406428×10^1505
0.50001: n~~6.3213988737×10^15051

Erratum: Limit[Log[xn]/(2 Log[n/2])... in the solution should read  Limit[Log[n]/(2 Log[n/2])...

Edited on September 29, 2011, 10:05 am

Posted by broll on 2011-09-27 23:14:21