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 Power 2 Sum Power 3 And 5 (Posted on 2008-10-04)
Determine all possible pair(s) of nonnegative integers (P, Q), that satisfy this equation:

2P = 3Q + 5

 See The Solution Submitted by K Sengupta No Rating

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 re: asking Charlie -- more detail | Comment 7 of 11 |

More detail on what UBASIC can do:

The following short UBASIC program tests its accuracy and capacity:

10   P1=1
20   for I=1 to 1000000
30     P1=P1*2:P2=2^I
40     if P1<>P2 then stop
50     print I:print P1:print P2
90   next

It ended on an overflow trying to do 2^8672. Up through 8671, there was no mismatch between 2^i calculated by repeat multiplications and that calculated by the exponential symbol.

2^8671, the highest it handled successfully, has 2611 decimal digits:

1702520705564891483448834809046454398393476719417318842658188433103599686824893
92715906718467007614597560460320859036854597205767388686312347414930801747971011
00327221992169069053388387454841282673389299897434427776534708187700985970817872
88801999724994864655431887433029440868853388125885208994867766100461802119118040
68253703073860896441205021509257452513465787299856436303129025474408791970772500
37798569515997141363135352925937330238536278846203603899597688863749134164028415
07241986299668227305345213106190914928913633629464465018974249197073367281945541
30269612814611254174394554030696720236853395034335068351638478715075963385148623
87948157980688856172199642692286837436104826054381069067991739425166696200873869
57664300403324421634165071565348972340088796847706134634703494097922425305020597
08187741146539749096608802788678243110043937548164863083693820871122168808109137
43471231425467749835476723429192426798746394464262761604199103998308990463786174
48296522276074666746236560898416975086246190364103562189550882148434345364214245
68466660760244469066720627719259792077771387940655787071426694759721128966234886
85404061301596243806332102413597321536817653818984461764363218367749250154672291
77113292018257153793234497829749228280455454624981960059426991001107256885048452
82895145154152000446631891216197002913110467243251201069421636940331549104985480
71878942235122987548210685455352075657934544876832881897601651938838019513410726
52657769600989623467536076562480620625274404640312983473656995180087493954473058
26590041936574855177636978962974165117165322018474710016624971812052988919747261
20440805027283260850964212170643924327433628254724298012279994430368173808039802
00344542796739890408995176906397017285182661556588202001837498581341214408098083
70159618778224074188377763784602608091564681347386134946301495999542742314159716
37341039793161938242258922733273926504863013718125446793283390732673019564623482
10895670680733655717993201781162858522555748034777543093513407918996211955240594
53640011456920250742833578165626504449422843268699025757614584976964386374491557
88986136420851818336726407277561087365696938467376439224550548172209095719098699
01270123335021665479384918117275251992774494254051196299052421868998443709372169
22045914688755531142524994207062314473151580759991321985870276475657753414393296
72025690619765747033035657973701273189799252847535931773759827175805491288647406
92391170042576394504018580051925678693001344559299247310323114791551720311150391
29995274147779648318267610902773645986578603365259394264181559538402818018976866
6488461955723831169315463395800403646568646047694848

By the way, 2^1000 has 302 digits, not 301, as log(2^1000) = 301.02999566398 (base ten), so 2^1000 ~= 1.071508607186 x 10^301, which has 302 digits.

?2^1000
1071508607186267320948425049060001810561404811705533607443750388370351051124936
12249319837881569585812759467291755314682518714528569231404359845775746985748039
34567774824230985421074605062371141877954182153046474983581941267398767559165543
946077062914571196477686542167660429831652624386837205668069376

 Posted by Charlie on 2008-10-05 12:51:05

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