What are the last two digits of 3^2002? Thanks!?

Question

How did you obtain 82? I got 09.

Hi, can you pls tell we where did you get the whole list of figures? Thanks!

Answer 1

My orginal answer of 82 is not correct. The correct answer is 09. Here's how I approached the problem.

2002 = 2*7*11*13

I rewrote the problem of 3^2002 as the following:

((((3^2)^7)^11)^13)

I then computed the last 2 digits of each power.

For example:

(3)^2 = 9
(9)^7 = 4782969 - for the next power computation only use the last two digits
(69)^11 = 168787390185178426269
(69)^13 = 803596764671634487466709

I chose to use the last two digits because as you can see the numbers get very very large.

So the answer is 09.

Answer 2

So here's a neat fact for you: If n is a number which is not divisible by 2 or 5, then the last two digits of n^40 are "01"! This is a specific case of "Euler's Theorem". It works for 3, so the last two digits of 3^40 are "01". Since 2000=40*500, the last digits of 3^2000 are "01" as well. Finally multiply by 9 to get that the last two digits of 3^2002 are "09".

Answer 3

Canty,

Your question reads:

3 raised to the four-digit number 2002.

I continue to get "error" on my calculator.

I don't think there is a solution to this question.

Guido

Answer 4

1573084126550386448693977157248194513476156238691669
6660306992106296176396045479403623378228782129999680
4805562653074892989423159228795778276297278188427463
3685621142080979461443376665630488471167387228413353
9816534698011930504505830325422370331317148623449356
8832416128143169592324271547018677959812487130934029
2696365781765886302065177796657337820571775423776216
2266819948999936197719315363278132816797971379421611
3978252953190018084218575761601073074575727809810142
9363891324372362915979982078767162066084341140124193
4824848013192441972209197252647251662359699706179955
6516070516574169914283327807510431880995658446637678
6836641458133292116438835664027483051516211782251215
3538833217317799745366768242767489432362347698506418
3976786985585702044106664004187269457210736342389154
1172740050968744147862815132889023166942708321381492
8407095589378985134200424423373801837708241471796217
5055479098786107320051034227936553925084209390120499
25064371486993960009

So it looks like 09 are the last two digits

Answer 5

it cant end with 09 ..it should end wid 3 ..........and da last two digits are 63

Answer 6

this is 3^2002:
15730841265503864486939771572481945134761562386916696660306992106296176396045479
40362337822878212999968048055626530748929894231592287957782762972781884274633685
62114208097946144337666563048847116738722841335398165346980119305045058303254223
70331317148623449356883241612814316959232427154701867795981248713093402926963657
81765886302065177796657337820571775423776216226681994899993619771931536327813281
67979713794216113978252953190018084218575761601073074575727809810142936389132437
23629159799820787671620660843411401241934824848013192441972209197252647251662359
69970617995565160705165741699142833278075104318809956584466376786836641458133292
11643883566402748305151621178225121535388332173177997453667682427674894323623476
98506418397678698558570204410666400418726945721073634238915411727400509687441478
62815132889023166942708321381492840709558937898513420042442337380183770824147179
6217505547909878610732005103422793655392508420939012049925064371486993960009

find the last two digits.