Hello, I was randomly browsing sites and I came upon this math question which I am stuck on (and desperately want to solve!): Find the last two digits of 3^3^3^3. Can you find the last three digits?

I'm stuck on this problem because I really don't know where to begin. I figure that you need to find a pattern or period of some sort, but the numbers grow too fast and therefore there aren't that many readily available samples. Any help is appreciated! THANKS!
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 Recognitions: Gold Member Science Advisor Staff Emeritus You can figure it out using modular arithmetic. a==b (mod n) means a-b is divisible by n. To figure out the last digit, use (3^3^3^3)==last digit (mod 10). Then use the properties of congruences to reduce 3^3^3^3 down to a single digit number. Similarly, the residue mod 100 leaves you with the last 2 digits. I'm sure there's info on modular arithmentic and congruences on Wolfram. If not, try Googling "modular arithmetic" and/or "congruence".
 Recognitions: Gold Member Science Advisor Staff Emeritus For instance, (all of the following is mod 100 ) 3^3=27 3^3^3=27^3=27*27^2=27*(25+2)^2==27*(25+0+4)=27*29=28^2 - 1=(30-2)^2 - 1== 0 - 20 + 4 -1 == -17 So, 3^3^3^3=(-17)^3== (-17)* 89 == (-17)*(-11) ==87 That should be your last 2 digits. If you want the last 3, you do the same thing mod 1000, instead of 100. If there's an easier way, I haven't figured one out yet.

Oh ok. I'm not really familiar with modular arithmetic but thanks!
 Recognitions: Homework Help Science Advisor 3^3^3^3 I'm going to assume that this is ((3^3)^3)^3 since I've seen it before (even though it goes against PEDMAS). 3^3 = 3*3*3= 27 ignore the second digit since it won't affect the first in Z multiplications 7^3 = 343 same reasoning 3^3 = 27 the last digit is 7 3^3 = 3*3*3 = 27 27^3 = 19 683 ignore all digits except first two since they won't affect them in Z multiplications 83^3 = 571 787 The last two digits are 8 and 7. I could do it again for three digits if you want.
 Recognitions: Gold Member We look at: 3^10==49 Mod 1000, 3^20==401 Mod 1000, 3^50==249 Mod 1000, 3^100==1 Mod 1000, and 3^7==187 Mod 1000. Then we have, using the usual notation 3^3^3^3 =3^3^27= 3^((3^10)x3^10x(3^7))==3^(401x187) ==3^(987) ==(3^900)(3^87) ==3^87==((3^20)^4)(3^7) ==((401)^4)(3^7) == 601x187==387 Mod(1000). Thus the last three digits are 387.