1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Last two digits of 16^198

  1. Jul 24, 2010 #1
    1. The problem statement, all variables and given/known data

    What are the last two digits of 16^198

    2. Relevant equations



    3. The attempt at a solution

    I begin by looking for a pattern.

    16^2 = 256
    16^4 = ...36
    16^6 = ...16
    16^8 = ...96
    16^10 = ...56

    so i see that the pattern repeats itself every 4 times. Then, i grouped the power as follows,

    (2,4,6,8) , (10,12,14,16) , (18,20,22,24) , ... , (194,196,198,200)

    The last number of each group has formula of 8n and that's how i got the last group.

    198 is third in the group which corresponds to the number having 16 as its last two digits.
    But i think i went wrong somewhere because the answer given is 96. Am i totally off track?
     
  2. jcsd
  3. Jul 25, 2010 #2

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    Re: indices

    You missed one.

    1610=1099511627776

    It's 1612 that ends in 56.
     
  4. Jul 25, 2010 #3
    Re: indices

    Thanks Vela, i am using a 570 calculator and it gives 1.099511628 x 10^12 for 16^10 so how do i know its last two digits?
     
  5. Jul 25, 2010 #4

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    Re: indices

    Just keep the last two digits in your calculations. The higher order digits don't matter.
     
  6. Jul 25, 2010 #5
    Re: indices

    yes, i know. The problem is how do i know the last two digits from 1.099511628 x 10^12. It looks to be 00. That's the maximum degree of accuracy my calculator can measure to.
     
  7. Jul 25, 2010 #6
    Re: indices

    Think about this... u'd get the last two digits of 16^198 as a remainder if u divide it by 100. Correct?

    So, u have to find 16^198 (mod 100). To do this you can use, general theorem

    a*b (mod c) = a(mod c) * b(mod c).
     
  8. Jul 25, 2010 #7
    Re: indices

    if you want to use ur calculator for solving this, but can't because of limitations in accuracy, here is what u can do.

    16^10 = 1.099511628... X 10^12 is clearly an approximation. To find the last two digits, subtract from this approximate number the exact number 1.099511628 X 10^24. This will give you the remaining digits of the answer. If that is too long again, repeat. Finally, you should get the last two digits.

    However, I suggest the method given above by me. It is easier.
     
  9. Jul 25, 2010 #8
    Re: indices

    thanks i figured out another way of getting the last two digits.

    Btw, i am not familiar with modular arithmetic but will pick up very soon.
     
  10. Jul 25, 2010 #9

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    Re: indices

    You can toss any higher digits even on intermediate calculations.

    162 = 256 ⇒ 56
    164 ⇒ 56*56 = 3136 ⇒ 36
    166 ⇒ 56*36 = 2016 ⇒ 16
    168 ⇒ 56*16 = 896 ⇒ 96
    1610 ⇒ 56*96 = 5376 ⇒ 76
    1612 ⇒ 56*76 = 4256 ⇒ 56

    Even if you don't know modular arithmetic, this should make sense to you. Any digit in the 100s position or higher can only contribute some multiple of 100 to the product, so it won't affect the two least significant digits. For example, consider 256*256. We can write

    2562=(200+56)(200+56) = 2002 + 2(200)(56) + 562

    The first two terms will be some multiple of 200, so their bottom two digits have to be zero. Only the last term, 56*56, contributes to the two lowest-order digits.
     
  11. Jul 25, 2010 #10
    Re: indices

    yeah thanks, i figured that out.
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook