digits 1000
vpa(10)^99 / (vpa(10)^31 + 3)
finds the part before the decimal point is
99999999999999999999999999999970000000000000000000000000000008999999
making the last two digits of the floor 99.
The fractional part, that was dropped by the floor process, comes out, to the given precision, as
.9999999999999999999999973000000000000000000000000000008099999
99999999999999999999999757000000000000000000000000000072899999
99999999999999999999997813000000000000000000000000000656099999
99999999999999999999980317000000000000000000000000005904899999
99999999999999999999822853000000000000000000000000053144099999
99999999999999999998405677000000000000000000000000478296899999
99999999999999999985651093000000000000000000000004304672099999
99999999999999999870859837000000000000000000000038742048899999
99999999999999998837738533000000000000000000000348678440099999
99999999999999989539646797000000000000000000003138105960899999
99999999999999905856821173000000000000000000028242953648099999
99999999999999152711390557000000000000000000254186582832899999
99999999999992374402515013000000000000000002287679245496099999
99999999999931369622635117000000000000000020589113209464899999
999999999993823266037160530000000000000001853020188851841
This does make an interesting pattern in itself.
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Posted by Charlie
on 2023-11-18 10:09:12 |
computer solution
