<p>I stumbled upon this interesting math question :-</p>
<p>"Without using a calculator, find the last 2 digits of 11^2006"</p>
<p>Can anyone help me out?</p>
<p>61- ten characters</p>
<p>the last two digits should be 61</p>
<p>proof by extension:</p>
<p>11^1 = 1
11^2 = 121
11^3 = 1331
...
11^2006 = .........61</p>
<p>QED</p>
<p>A more elegant proof :)</p>
<p>11^1 = 11
If we assume,
11^k =
.k1
Then, the following must be true
11^k+1 =
(k+1)1</p>
<p>By def. 11^(k+1) = 11^k + 11^1 = ..(k+1) --- Q.E.D</p>
<p>Therefore, 11^i = ......i1 for all * i E Pos. Int *</p>
<p>Rabban:</p>
<p>That's like the Gauss proof for sums of consecutive integrals...know for 1, assume for n, proove for n+1, induction, I think it's called?</p>
<p>I like proof by induction, seems more airtight.</p>
<p>A noninductive proof:
11^n = (10+1)^n
= 10^n + C(n,1)*10^(n-1) + ... + C(n,n-1) * 10 + 1
= C(n,n-1) * 10 + 1 (mod 100) = 10n + 1
So the last two digits ba are b = n (mod 10), and a = 1.</p>
<p>[edit] oops......</p>
<p>108075294440392931176004076072085611902822482342104534854446080512585760361171
782543232489504237035009325910669383458045377511386080915746345242851605663933
822767141684768341591140537330359564694533418996840017322297413869438464056768
449699782585926787003242269844359101567354575831189953867859653050452189490651
935625672323692103990619402222592286547054402099473239757254052391076595485345
545430458343566916147435907365458425513333528620813262921746044956150110215917
019496276759656211010961284251928080881227236887866203463309123412145974296836
950499803834936777737263763354458002341208261152775605887939014144739608885075
162813350607325137690524869817397440170308097045069449357089277131794164755313
000346811069374734224259484733909407377943401541056248309845242763016413103137
093357037285352413059267400058380640147682523055132609124421934119766737082532
657291211134568613396856268104562530081177557726390528890580884961215554659869
589073643609882695576449465157679447000609297850200276030308778566857701285868
186719485927849559299248972168755528630886989633436834720953247386129756677860
993968073274340253999377381807885179411398143900314188860360922891560899321544
613847801530918920466902949753028329535329114047005912253329204144252707877475
142780010737566959219067426048040348353382077858409252704341492768521599841872
974095771625867565178896950717082496598889620376778647135508781769164391061326
494917446532814105346238598462853155179275778871673592509132985719729186838300
529881031512198559981189820777763147392210578895300443242014541602615756926170
816945891195351540528144179487972589759666714873041013812248710317976944378647
999262888587801771811545564685108267885651081361474063041615307568400246393879
044981009146939788620339084090945624125516488801686491512397323143820830682362
066726934124787456366866066914140878071511533956049335230559557910240633686016
569976420611570721940013581648032055669966809230479673392068894834218862341568
517990415907734655580026981008586438520456612441102296582651720604522402667399
14205854952437469054019352649625798932333445119987400485891561</p>
<p>so...61.</p>
<p>Mathematica isn't a calculator...;-)</p>
<p>By the way, you will notice that the powers of 11 fall apart into pascal's triangle</p>
<p>So, 11^5 quickly? just find the 5th row, 1 5 10 10 5 1
carry the 10's:</p>
<p>1 6 1 0 5 1</p>
<p>wooah.
thanks</p>
<p>You can use the mod system to solve it, but I'm not sure you want to learn how to do all that number theory stuff.</p>
<p>I just used modulus for the sake of formality, it's clear that only the last two terms in the expansion of (10 + 1)^n affect the answer.
The non inductive proof I gave better expains the phenomenon, since it is clear that if you expanded (10+1)<em>(10+1)</em>...n times...*(10+1), realize that each of the 2^n terms in the binomial expansion (without having combined like terms yet) chooses either 10 or 1, n times; only the cases where 10 is chosen one or fewer times applies - and there are n ways to choose 10 once and 1 way to choose 10 zero times...</p>
<p>So it's not number theory, it's just combinatorics</p>
<p>I must have missed your post. You're definitely right, it is combinatorics. I just grouped it into number theory, since I learned it in a number theory unit. :)</p>
<p>11^k will give the pascal's triangle at k</p>
<p>Ex. 11^2=121
11^3=1331</p>
<p>So for 11^2006 it has to end in one and the previous term will be 2006 thus second to last number is the 6 in 2006</p>