The sum of the reciprocal of the square root of all the positive integers up to n is denoted by F(n), that is:
F(n) = 1+1/√2 + 1/√3 +...+ 1/√n
Determine the maximum value of n such that the integer part of the base ten expansion of F(n) DOES NOT exceed 2012.
*** For an extra challenge, solve this puzzle without using a computer program.
Maximum such n is 1,013,505:
10 point 35
20 while T<=2012
30 PrevT=T
40 N=N+1:T=T+1/sqrt(N)
50 wend
60 print N-1,PrevT
70 print N,T
finds that F(1013505) ~= 2011.9998512 and F(1013506) ~= 2012.0008445.
Specifically, the output is:
Words for fractionals 35 (Decimals for display 168)
1013505 2011.9998512067347756246902699409366862982247383561279030457342
16555503796507039814234377477881798284087714660672885032442019561678422057764197
776286526187964973375524958049
1013506 2012.0008445213783854834399657352730035024527421198563401973979
52251518173653291207749387862141705733740828106568073177121727135534711192409921
224060824796931311757117336879
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Posted by Charlie
on 2012-03-16 12:46:14 |