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Power Tower Help Please.

  1. May 11, 2004 #1
    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. jcsd
  3. May 11, 2004 #2

    Gokul43201

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    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".
     
  4. May 11, 2004 #3

    Gokul43201

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    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.
     
  5. May 11, 2004 #4
    Oh ok. I'm not really familiar with modular arithmetic but thanks!
     
  6. Jun 29, 2004 #5

    Alkatran

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    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.
     
  7. Aug 5, 2004 #6
    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.
     
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