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 Summing inverses II (Posted on 2012-03-16)
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.

 No Solution Yet Submitted by K Sengupta No Rating

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 computer solution | Comment 2 of 12 |

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

 Posted by Charlie on 2012-03-16 12:46:14

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