Power Tower Help Please.

ilml

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!

Gokul43201

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".

Gokul43201

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.

ilml

Oh ok. I'm not really familiar with modular arithmetic but thanks!

Alkatran

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.

robert Ihnot

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.