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Math Help - units/ones digit

  1. #1
    Senior Member sfspitfire23's Avatar
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    units/ones digit

    Hi all,

    if given a number like 56^{34}, how would I find the units digit without multiplying the number out? What about for expressions such as (867)\times (563)\times (y-2) ?

    What is the method?
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  2. #2
    MHF Contributor undefined's Avatar
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    Quote Originally Posted by sfspitfire23 View Post
    Hi all,

    if given a number like 56^{34}, how would I find the units digit without multiplying the number out? What about for expressions such as (867)\times (563)\times (y-2) ?

    What is the method?
    We are interested in 56^34 (mod 10). Think mod 2 and mod 5. What do you notice?

    We are interested in (867)(563)(y-2) mod 10. (I assume y greater than or equal to 2. If y less than 2, still think mod 10 but make appropriate adjustment.) Reduce the first two factors mod 10. What do you notice?
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  3. #3
    Senior Member sfspitfire23's Avatar
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    Hm. So, 56^34 is equivalent to 6 (mod 10). So 6 is the remainder when 56^34 is divided by 10. thus 6 is the units place

    We have (7)(3)(y-2) (mod 10) So ones place is 1?
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  4. #4
    Senior Member sfspitfire23's Avatar
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    Why mod 10 though?
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  5. #5
    MHF Contributor undefined's Avatar
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    Quote Originally Posted by sfspitfire23 View Post
    Hm. So, 56^34 is equivalent to 6 (mod 10). So 6 is the remainder when 56^34 is divided by 10. thus 6 is the units place
    The conclusion is correct but you did not explain how you got there, so maybe you took a long way or made a lucky guess and I have no way of knowing. Here are listed many steps, but of course you can do it in your head without writing any steps.

    56^{34} \equiv 0^{34} \equiv 0 \pmod{2}

    56^{34} \equiv 1^{34} \equiv 1 \pmod{5}

    \implies 56^{34} \equiv 6 \pmod{10}

    Quote Originally Posted by sfspitfire23 View Post
    We have (7)(3)(y-2) (mod 10)
    Right, but keep going. What is 7*3 reduced mod 10?

    Quote Originally Posted by sfspitfire23 View Post
    So ones place is 1?
    How do you get this? This fails for many y, including y = 2.

    Quote Originally Posted by sfspitfire23 View Post
    Why mod 10 though?
    What do you think?
    Last edited by undefined; August 2nd 2010 at 11:02 AM. Reason: clarify wording
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  6. #6
    Senior Member sfspitfire23's Avatar
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    Did you find what made the expression equivalent to 0, then to 1, and multiplied them together?

    21(y-2) is equivalent to 1(y-8) (mod 10).
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    Hello, sfspitfire23!

    Find the units-digit of: . 56^{34}

    Did you crank out a few powers of 56?

    . . \begin{array}{ccc}<br />
56^1 &=& 5\boxed{\!6\!} \\ \\[-4mm]<br />
56^2 &=& 313\boxed{\!6\!} \\ \\[-4mm]<br />
56^3 &=& 175,\!61\boxed{\!6\!} \\ \\[-4mm]<br />
56^4 &=& 9,\!834,\!49\boxed{\!6\!} \\<br />
\vdots && \vdots \end{array}


    Do you see a pattern?
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  8. #8
    MHF Contributor undefined's Avatar
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    Quote Originally Posted by sfspitfire23 View Post
    Did you find what made the expression equivalent to 0, then to 1, and multiplied them together?
    I don't know what you mean. First of all, which expression are we talking about, 56^34 or (867)(563)(y-2)?

    For 56^34, I use the Chinese remainder theorem. The algorithm is unnecessary because the numbers are so small; just ask, what even number is there belonging to {0,...,9} that is congruent to 1 (mod 5)? Or, consider all numbers in {0,...,9} congruent to 1 (mod 5), which are {1,6}, and take the even number.

    Quote Originally Posted by sfspitfire23 View Post
    21(y-2) is equivalent to 1(y-8) (mod 10).
    Why did you change y-2 to y-8? It is not equivalent.

    We have 21(y-2) is congruent to 1(y-2) which is just y-2, so one way to proceed is to choose y' congruent to y (mod 10) such that y' is in the set {2,...,11}, then the units digit is y' - 2. Another way is to take the common residue of y, call it for example z, then if z < 2 take z+8, otherwise take z-2. (That is assuming y greater than or equal to 2.)

    Additional note: To formalise Soroban's post, we have 6^2 \equiv 6 \pmod{10} so the last digit must be 6; this is a bit faster and simpler than what I recommended, but either way with experience you can get the answer in not more than a few seconds.
    Last edited by undefined; August 2nd 2010 at 11:42 AM.
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  9. #9
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    Quote Originally Posted by sfspitfire23 View Post
    Why mod 10 though?
    This may answer your question in general...

    Notice that every positive integer N can be represented uniquely in the follwing way: N=c_n10^n+c_{n-1}10^{n-1}+\cdots +c_110^1+c_0, where 0\leq c_i< 10 ( i=0,1,...,n) and c_n\neq 0. Here, the c_i are called the digits, and the expansion c_n10^n+c_{n-1}10^{n-1}+\cdots +c_110^1+c_0 is written as c_nc_{n-1}... c_1c_0 in short.

    For example, 531=5\cdot10^2+3\cdot10+1.

    Because c_i10^i\equiv 0(mod\ 10) for i>0, we have N\equiv c_0(mod\ 10). This means that if you compute a positive integer modulo 10 and take the least nonnegative residue you find the last digit, c_0.
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  10. #10
    MHF Contributor undefined's Avatar
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    Quote Originally Posted by melese View Post
    This may answer your question in general...

    Notice that every positive integer N can be represented uniquely in the follwing way: N=c_n10^n+c_{n-1}10^{n-1}+\cdots +c_110^1+c_0, where 0\leq c_i< 10 ( i=0,1,...,n) and c_n\neq 0. Here, the c_i are called the digits, and the expansion c_n10^n+c_{n-1}10^{n-1}+\cdots +c_110^1+c_0 is written as c_nc_{n-1}... c_1c_0 in short.

    For example, 531=5\cdot10^2+3\cdot10+1.

    Because c_i10^i\equiv 0(mod\ 10) for i>0, we have N\equiv c_0(mod\ 10). This means that if you compute a positive integer modulo 10 and take the least nonnegative residue you find the last digit, c_0.
    A bit simpler is to just lump the c_i with i > 0 together, as in 12345 = 1234 * 10 + 5.
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