All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars
 perplexus dot info

 The Unknown Digits (Posted on 2006-04-18)
Determine the last six digits of the decimal number given by 7^(3^3002).

 See The Solution Submitted by K Sengupta No Rating

Comments: ( Back to comment list | You must be logged in to post comments.)
 solution | Comment 1 of 2

The extended precision of UBASIC is of use here:

The last two digits of powers of 7 repeat in a cycle of 4:
07, 49, 43, 01

The last three digits repeat in a cycle of 20, as in the following table of the last 6 digits of these powers of 7:

` 21          284007 41          568007 61          852007 81          136007101          420007121          704007141          988007161          272007181          556007201          840007`

the program being:

10   for I=1 to 40*20 step 20
20   print using(8,0),I,(7^I)@1000000
30   next

From the above table, you can see that the last 4 digits repeat in a cycle of 5 * 20 = 100, so we tabulate for every 100 powers of 7:

`    1               7  101          420007  201          840007  301          260007  401          680007  501          100007  601          520007  701          940007  801          360007  901          780007 1001          200007 1101          620007 1201           40007 1301          460007 1401          880007 1501          300007  `

From this we see the last 5 digits repeat in a cycle of 5 * 100 = 500.

But we also notice that the sixth digit from the end is incremented by 1 for every repetition through the cycle of 500 terminal 5 digits, so 10 cycles of the last 5 digits will produce a repetition of the last 6 digits, so that is a cycle of 500 * 5 = 5000.

3^3002 is

`  2079728620300718345423798488066349561836471608956655431276203871987492001275962 90989154812574463616852290074576255353905354265133756437274517818248266136217643 56693788687777156060717839632488676063160943502334938195902355590039508301313069 00484892548305503906527600244921860564267819848672643008121931279712424696662433 90939948603503931985960890214895835288218744967738832144124766633498119800658307 22002954793923799569704901966583073300996164099418118584636877069394057598294248 13407836154017923699626628159946224932862275102842814118137581960255160611640453 96469486949288152050067178518591163815911680886987427011833813986700647016729349 48621180176650597749108280988742836198992828871786569998239204915151952573462040 80558070339442748583606309836467205614456720578377583830989982229040632017444595 04687285424957155058461739091217482769490690052807269879218870540734718594886443 96615217020591290685465755569603402429618387082668422656202519528893479645848824 51145171094521832921632693375060369253706719349376957476192658300763616503966830 89995784213848930358516776731175596309946662121602576765698018688032885248861429 00074813662311685180817679746487657489858651256268683144008352382682993538953354 63629929973858821717470208967495676929019257936278538066429572037407491929887738 67656479870881694128504272901859288632587695067456545336246109089059378918071967 83890108540085333497949726248859152147330229354934347815928186083890940009  `

which is congruent to 9 mod 5000.

(The same could be done on a calculator with less precision by repeated multiplication by 3 and taking mod 5000 along the way; plus some shortcut powers by factoring 3002 into 2*19*79)

So the last 6 digits of the number sought is the same as the last 6 digits of 7^9.
Since 7^9 is 40,353,607, those last 6 digits are 353607.

 Posted by Charlie on 2006-04-18 14:01:35

 Search: Search body:
Forums (0)