1. May 11, 2004

### 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!

2. May 11, 2004

### Gokul43201

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".

3. May 11, 2004

### Gokul43201

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.

4. May 11, 2004

### ilml

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

5. Jun 29, 2004

### 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.

6. Aug 5, 2004

### 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.