Midterm Practice: Find the last digit of a number

  1. What is the last digit of [tex]2222^{50}~+~7777^{16}[/tex]?

    I know how to do these kinds of questions, but this time around I decided to try mod 11. Is there a reason why mod 11 does not work in this case?

    [tex]2222 \equiv 0~(mod~11)[/tex]
    Therefore, [tex]2222^{50} \equiv 0^{50} (mod 11)[/tex], that is
    [tex]2222^{50} \equiv 0 (mod 11)[/tex]

    I did the same for 7777. [tex]7777^{16} \equiv 0 (mod 11)[/tex]

    Thus, I concluded that [tex]2222^{50}~+~7777^{16} \equiv[/tex] 0 + 0 (mod 11). And hence, the last digit is 0.

    The last digit is actually 5.

  2. jcsd
  3. The logic you are using will imply that 2222^anynumber + 7777^anynumber will always produce the last number to be zero. One Example would be 2222+7777= 9999. Anyway, what makes you think that mod 11 will work?
  4. I just thought you could use mod n for any integer n. I didn't think there was some formal way of doing it. So I am assuming now there is, and that would be mod 10 correct?
  5. Take a number, say 123. Notice that division by 10 will produce the number 12.3, that is to say, 123/10 = 12 + 3/10 <=> 123 = 12*10 + 3.

    Do you see now why reducing a number modulo 10 will produce the last digit of that number?

    Or you could notice that a number with decimal expansion "x_n ... x_1 x_0" is equal to x_0 + 10x_1 + ... + 10^n * x_n. Taking that modulo 10, we see that every term except x_0 disappears, i.e. "x_n ... x_1 x_0" == x_0 (mod 10).
  6. Yea, I can see why 10 does it so well :D I wasn't sure that mod 10 was the only way to do it. Thanks for clarifying that.

    The reason I tried mod 11 is because when I tried mod 10, I had a remainder of 2. And when I tried to find a pattern in 2^n from n = 0, the pattern repeats every four after 2^0. The pattern is: 1,2,4,8,6,2,4,8,6... and so forth. I don't know what to do next because of that 1 in the pattern. Normally, I would divide the length of the pattern into the exponent, but I'm not sure if that would work as there is an exception.

    Could you or anyone explain to me how I would find the last digit for 2222^50?
  7. *edit* I really should read the whole post before posting.

    But if you start from n = 1 instead, you get 2, 4, 8, 6, 2, ... . That is, every number of the form 2^n with n a multiple of 4 ends with 6, every number for which n is "a multiple of 4" + 1 ends with a 2, etc.
    Last edited: May 27, 2005
  8. Ahh okay, so I guess I needed to look at the pattern more closely.

    So I wouldn't be dividing the exponent by the length of the pattern then... right? All I need to do is figure out what it is a multiple of and apply that observation you made.
  9. Do you see why for both powers of 2 and powers of 7 you have to look at the value of the exponent mod 4 to get the last digit of the number. You need not factor the exponent, just divide by 4 and look at the remainder.
  10. Yea, thats what I would normally do. My problem is regarding the exception in the pattern. Notice that 2^0 = 1. And that is the only time in the pattern where you will see a 1. I wanted to know how to deal with such cases. I ended up discarding 2^0.

    Can anyone tell me a rule of when such situations arise?
    Last edited: May 28, 2005
  11. There is no problem with the exception of n^0 or for any other exceptions. For even M values the 1 can not form a part of the repeating cycle with n being even since an even number mod an even number is always even. All that is necessary is to calculate the indicies, for n^0, n^1, n^2, n^3 ... until you get a repetition. This may or may not include the 1 in the repeating cycle, but once you get the repetition in the series the true cycle becomes immediately apparent. If the 1 is not included and the cycle is K terms and if n^C mod M is the first value repeated then you can say:
    n^(A mod K) = n^(A) mod M for A greater or equal to C.
  12. By using mod 11 you've shown it ends in 0 in base 11 (it ends in 5 in base 10).
  13. Borek

    Staff: Mentor

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