stekas@hou2g.UUCP (J.STEKAS) (02/13/84)
Complete the series.
110, 20, 12, 11, 10, ...
Jimamigo2@ihuxq.UUCP (John Hobson) (02/13/84)
>> Complete the series. >> >> 110, 20, 12, 11, 10, ... Simple, the answer is 2, 1, 0. What we have here is the series 7, 6, 5, 4, 3 ... in base 3. John Hobson AT&T Bell Labs Naperville, IL (312) 979-0193 ihnp4!ihuxq!amigo2
amigo2@ihuxq.UUCP (John Hobson) (02/14/84)
In my answer to the problem: >>>> Complete the series. >>>> >>>> 110, 20, 12, 11, 10, ... >> >> Simple, the answer is 2, 1, 0. What we have here is the series >> 7, 6, 5, 4, 3 ... in base 3. I made an error. 110 is not 7 base 3, but rather 12 base 3. If the original proposer of the problem says that 110 is a typo for 101, then I will allow my answer to stand. If it is not, then I withdraw it. (My thanks to Don Stanwyck for pointing this out to me.) John Hobson AT&T Bell Labs Naperville, IL (312) 979-0193 ihnp4!ihuxq!amigo2
john@genrad.UUCP (John Nelson) (02/14/84)
>> Complete the series. >> >> 110, 20, 12, 11, 10, ... Answer: 6, 6, 6 ... What we REALLY have here is the number six in bases 2, 3, 4, 5, 6 ... Once base 7 is reached, then the representation never changes.
ags@pucc-i (Seaman) (02/14/84)
The completion of the series 110, 20, 12, 11, 10 , ... is "an infinite number of 6's". The reason (rot13): Gur Agu grez bs gur frevrf vf fvk rkcerffrq va onfr (A cyhf bar). -- Dave Seaman ..!pur-ee!pucc-i:ags "Against people who give vent to their loquacity by extraneous bombastic circumlocution."
kaufman@uiucdcs.UUCP (kaufman ) (02/15/84)
#R:hou2g:-16700:uiucdcs:40700005:000:272
uiucdcs!kaufman Feb 14 11:07:00 1984
101 is still not 7 base 3 (21 is). Anyway, here's a working solution (rot 13).
Rirel sbyybjvat grez vf n fvk (nq vasvavgrz). Gur frdhrapr vf jevgr gur ahzore
fvk va onfr gjb, onfr guerr, onfr sbhe, ...
Xra Xnhszna (hvhpqpf!xnhszna)csc@watmath.UUCP (Computer Sci Club) (02/16/84)
110, 20, 12, 11, 10, 6, 6, 6, 6, ...
William Hughes