What are the last two digits of 2017^{2017}?

[param-97]

yes I do , but i cant use calculator or technology

390657975543372574454980188159622730316111397478633174280801431635240420786678101177943257944177583985693036750913176864857698546600616339008210220734937908444842063745237656850478885929417325679578963373568227616903361285861720928485319957896879451656212960334524148908182859288925011902051241977461866923828580514121684519464635969277393017389045213030148410860370490766318591802799658830628904573310183798372803277414275857819439349070663831157336029322681657747930292509484531407316245237233223535287180520182692120301190715379956212549955472924854051368917564413054345164976023589951663964251322964845660608570604879616601321177653122329485885877061828720295615315907090902544524461292926690743540573405820595406619662579444139540489388779647728869585635479195842669551613833202622561396245073359759987677556858143344727668109856905808726786650150176346749276701349195925308447304906625583450532575618820393030470662264833655851676447780085570224770196613311539211965922904482259190533327063104392766603752817849162919553856846462339898292286551155249369946391847509120475648953747207646721406954090955923631733324724595764624045478715399572715756795527704408680018385825873357580378471551915354156980027304848137549295408442862472069365322824672593499952640854410304521328639841469979889738057756391887864581212586380773282199133210786927259058491740244495119211462941902629069960853689596035764784053231433840803883793195549088565567315908088452296187414871507326380105914169781913896257814769011618619675396354441317274069735446648347083968532247927971561024407596290964720192472655232577086691267038690883293706899164421798556145451861350077011342811377914710690611991955035276929415307308068682441254479160294779193653509071602125556005926007272962151491655468415250273052830341020248623946192353295309806703225240777982731701762090578986703739265444559438301724941322343380259142075186404633958296871694661388388363008928909285016674242009847561523172702925696786989760986029187450917382699380538627161034744426951667256700015439675804055100094410479024451590607819278757744377853151071774880174452988351732344899456989727677340502177914005216463498939122497542108316809164832706303355325260893006805273079813392225356203349713572906588631985070013516668447331459865948614436309416691406526352511934470081928515538657796128428725229849486439078722383264002709446351625648515268700769721875710891517353990646067011044373315111199657801536917591352152223077856196758593774633883045958703352614309802218762142052453275488593184728802759403200439013018317636233481560894473946887064092502200732585614385533365417174868591302991871096647001079192734599003615765633603859333122333043348929219564612801024584337753852145099150440931873218035637881859414477740279483748996993268227511216215356580915170417907045718050420690639845019200539033361819456599440454575081778171829270612503157953281522607355349626354620104011641072405512170254572784586270506701505209700597593203542864418839562193815774326236977529972966261410337323664706113705568867451193992124787596100597886820452817195495502000070012798145376106754566805675636189183546179514300773274177825473054551165894880193962347892032287700827580232033226689816518092464826811804221423873093867168163180699559403340562561405808194971798503289109355115029124983991018380698424651876194056194282127149997079562496102454369536500984933908739935921807873526766938475347876733574757177344496981885814742371276010503960369796383469346470277399081631964686337475585520656880146671430375741270928466951119055375048418948497646736820963269750210123666066640911512953157476533586025093593927262144788949918692089397761431261871877904288178513169823048344238328226092694737357281270118671247017486305673456960008250271760371647008940741821145048849317096828419862967834904167454511084031404938107160029782440234575937475817925416626909085990876918495884680483366558853396587675219737345520051486671981990599704130000418965813007224733672096313749475119407724014818508929543517482095607287137608245178766538811447560502020386463181841378621278229474102676308905984144092314179309459435863896493726079330850497047869497570735068955181584557294719863535891289292988034828718933264617496210897466235050589604540196335236324754935438515933094258601062968745635884513724512140377812087996319489837611218393654491607009770616845923444210676366383537233648308641710278661628743001714662003390619173697170384492388045530484711672629944557695625184606039877731198679176733148119428829025189527784452428862612449080278884098946247489646956505581906115889044658676526114959262438720723867498045450888653408171059380160079360684890858305066549868679454778711971961004179359224085529666097900733242139003068648637827432609033040313477972019215266900544727175971758706967935502929482764524716791627240974091959226574012956618385918903583572362462995750127092270563921014720062585125823141782751812643109647530407871678226331249262326701055037286420326554368359401762702158677183942741326853694712155155415705784703907055557534253867338872894384930397539302092022885083011970919654037384692347973118459165661365055539195347218311488110187672218257397856026847785288284864740137044134948049741708836860375429551872331945923347901854407401857680985221609668998162351789574745008602677700153798722721150871343522597554976525812486209111636964612833470588367827664730339266308180436893671832778798527588104827613199838322557939018272344962449422866729130494318349786441889312965949010199609058366877816005182554357929793751308237195750885784710621366674313399801021398587793981696741636699303232640672376628017242474181779237945245393333912578179230125423667332678409969108298158976562935837830069836819609840129355006609312690665992668306149298048454144277912938192170665670364309688124177002766249617224562497248029025108484112804667962930792724655475851499391116160583635975886976803592692382140202756950689014310186759786818837653421528980068848505340581525138324224214436571777530620120022380670781205268816957660255816330863495713717498612056842009287332050038500758131655474444270464394270810368067423747143094562220685933183169518338219129062478901663665717169904302749323113722607056692170317367451511687434657711633713495885955145131153699809443084366039305972471814047033163613157827280690136693166560848826704613149166615379944905642268227182475956096597627288520629228771111807430370106408669790774313609604966005455290197434687532220849212300973638407916288237930524109849193392701173691235386179824246600514129898684575418268770378769936070701854151282858987245060220859376040188961430411125280662489310010991839503913438177

 

[pka]

I have flowed the replies to this question, And I would really like if someone here could tell me what in number theory one needs to know the find the last two digits in the expansion [imath]2017^{2017}?[/imath]
I gladly admit that I am woefully ignorant of number theory.
That said, if there is no basic theorem that tells us about trailing digits then I find this to be an irresponsible question.

 

[nanase]

I have flowed the replies to this question, And I would really like if someone here could tell me what in number theory one needs to know the find the last two digits in the expansion [imath]2017^{2017}?[/imath]
I gladly admit that I am woefully ignorant of number theory.
That said, if there is no basic theorem that tells us about trailing digits then I find this to be an irresponsible question.

That's why I am drawn to this question also, I wanna solve this question without using technology and using the simplest maths possible and available to my level. Mr BBB's approach is suitable for me.

 

[BigBeachBanana]

[math]2017^{2017} \equiv 17^{2017} \stackrel{*}{\equiv}17^{2017 \mod\phi(100)} \equiv 17^{2017 \mod 40} \equiv 17^{17}\equiv 17^{16} \times 17\stackrel{†}{\equiv} 81 \times 17 \equiv 77\mod 100[/math]
[imath]*\text{Euler's Theorem: } a^m \equiv a^{m \mod \phi(n)} \mod n, \text{where m and n are coprimes.} \quad [/imath]
[imath]\phi(n) \text{ is Euler's totient function.}[/imath]

[imath]†\text{Leveraging modular exponentiation to find } 17^{16} \mod 100 \text{ (as shown in post \#13).} [/imath]

 

[ddbingaman]

I need help in working this out
View attachment 36674
No other hint is given. What I tried is looking for pattern by typing it to the power of 1 , 2 and 3 and so on, but it gets too large and incomputable. What other way can I use and how? Thank you

2017^2017 on my hand held calculator gives me:
3.9065797554337257445498018815962e+6665

That is 2017^2017 notice it is in scientific notation and the number of zeros is 6665! A number so large it exceeds the number of atoms in the entire universe.

For the pattern is would go like this:

2017^0 * 2017^1 * 2017^2 (etc.)

Your can then factor out 2017 and it becomes:

2017*(0+1+2+3+4+5+6+7... all the way up to 2017)

Notice that number is going to be significantly larger than even 2017^2017.

Here is a spreadsheet that gives you an idea what it happening...



Hope that helps

 

[topsquark]

2017^2017 on my hand held calculator gives me:
3.9065797554337257445498018815962e+6665

That is 2017^2017 notice it is in scientific notation and the number of zeros is 6665! A number so large it exceeds the number of atoms in the entire universe.

For the pattern is would go like this:

2017^0 * 2017^1 * 2017^2 (etc.)

Your can then factor out 2017 and it becomes:

2017*(0+1+2+3+4+5+6+7... all the way up to 2017)

Notice that number is going to be significantly larger than even 2017^2017.

Here is a spreadsheet that gives you an idea what it happening...

View attachment 36722

Hope that helps

How does this have anything to do with the last two digits of [imath]2017^{2017}[/imath]?

Please keep it relevant.

-Dan

 

[ddbingaman]

How does this have anything to do with the last two digits of [imath]2017^{2017}[/imath]?

Please keep it relevant.

-Dan

True, if we just use the last two digits of 2017 that being 17 then
the math becomes:

17*(1+2+3+4+...17)

or

17 + 34 + 51 + 68 + 85 + 102 + 119 + 136 + 153 + 170 + 187 + 204 + 221 + 238 + 255 + 272 + 289 or

2,601 ways when you just use the first two digits.

 

[stapel]

True, if we just use the last two digits of 2017 that being 17 then
the math becomes:

17*(1+2+3+4+...17)

or

17 + 34 + 51 + 68 + 85 + 102 + 119 + 136 + 153 + 170 + 187 + 204 + 221 + 238 + 255 + 272 + 289 or

2,601 ways when you just use the first two digits.

That is *not* how exponentiation works.

 

[khansaheb]

True, if we just use the last two digits of 2017 that being 17 then
the math becomes:

17*(1+2+3+4+...17)

or

17 + 34 + 51 + 68 + 85 + 102 + 119 + 136 + 153 + 170 + 187 + 204 + 221 + 238 + 255 + 272 + 289 or

2,601 ways when you just use the first two digits.

The response above (#27) is so misguided that
- IT IS NOT EVEN WRONG.

 

[param-97]

I dig deeper into this problem. And I concludes that The solution to this problem is not to find the evaluation in any way but to know the unique property associated with this that seem to be complex in first appearance but its a simple problem with some basic knowledge of Even Numbers, Odd Numbers as well as Exponentiation. We will conclude the result with Number Parity and Exponentiation properties.

Observation 1 > Number Parity:
We can make some observations about the last two digits ( or any digits xyz ) of A Number ( Even or Odd) raised to the power of itself when the Number has N digits.

Let's denote this number as P with N digits.

A four-digit Number can be written in the form abcd, where a, b, c, and d are digits of that Number. Here the number of digits is N = 4. The last two digits of this number are cd.

Now, consider P ^P. The last two digits of P ^P will depend on the value of P. However, if we look at the last two digits of P itself, i.e., cd, we can make some general observations.

1. If d is even, then D ^D will end in 0, 2, 4, 6, or 8.
2. If d is odd, then D ^D will end in 1, 3, 7, or 9.

Therefore in Observation 1, the last two digits of P ^P will depend on the parity of the last digit d of the four-digit number P.
The Problem under consideration has all options which are Odd so we can't shortlist the options here.

Observation 2 > Exponentiation:
To calculate the x last digits of a power P ^P , you can use the modular exponentiation, but Evaluation is impractical as Number is very large here. So We use some special property of Exponentiation here:

Take this number as P with N digits again as a, b, c, and d digits. We will now write it as ABCD ^ABCD.
The Last Two Digits in complete exponentiation for any Number with N Digits (=4 below) has this below property:
ABCD ^ABCD = ABCD ^BCD = ABCD ^CD = BCD ^CD = CD ^CD

So, Now We have a very simplest form of exponentiation CD ^CD which is 17 ¹⁷...
We can now evaluate 17 ¹⁷ Mod 100 to find last two digits which will be equivalent to that of 2017 ²⁰¹⁷...

Let's begin with that. I recognized 17 ¹⁷ as being 1 greater than a power of 2. This suggests that one can proceed by squaring too which is the simplest case.

Now:
17 ² ≡ 89 mod 100
17 ⁴ ≡ 89 ² ≡ 21 mod 100
17 ⁸ ≡ 21 ² ≡ 41 mod 100
17 ¹⁶ ≡ 41 ² ≡ 81 mod 100

All that is left is to compute 17 ¹⁷ ≡ 17 * 81 mod 100
So is the answer, 77.

 

[pocobind]

I am not supposed to use calculator so this is probably not the right method also. But what am I seeing here? I don't see any pattern yet

 

[BigBeachBanana]

I am not supposed to use calculator so this is probably not the right method also. But what am I seeing here? I don't see any pattern yet

See post#24.

 

[Steven G]

2017^2017 on my hand held calculator gives me:
3.9065797554337257445498018815962e+6665

That is 2017^2017 notice it is in scientific notation and the number of zeros is 6665! A number so large it exceeds the number of atoms in the entire universe.

I will work with the answer you got from your calculator and your comment. Where do you think those 6665 zeros go? I would think that you would say that they go at the end. So why not conclude that the last digits are 00.

Your comment about 6665 zeros is wrong! The e+6665 means that the decimal goes to the right from where it is (!!) 6665 places. There are about 30 digits that are listed to the right of the decimal which are part of the 6665. That is, there are 6665-30(about) digits that are not listed. They are not necessarily all zeros!!!

BBB gave you the method to solve your problem in post #24

 

[Agent Smith]

[imath]7^1 = 7[/imath]
[imath]7^2 = 49[/imath]
[imath]7^3 = xx3[/imath]
[imath]7^4 = xx1[/imath]
[imath]7^5 = xx7[/imath]
[imath]7^6 = xx9[/imath]
[imath]7^7= xx3[/imath]
[imath]7^8 = xx1[/imath]


[imath]7^{2017} = 7^{4 \times 504 + 1}[/imath]

Is the last digit 7?

If si, answer = 57 or 77