Hi all,

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

What is the method?

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- Aug 2nd 2010, 10:30 AMsfspitfire23units/ones digit
Hi all,

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

What is the method? - Aug 2nd 2010, 10:41 AMundefined
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? - Aug 2nd 2010, 10:45 AMsfspitfire23
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? - Aug 2nd 2010, 10:48 AMsfspitfire23
Why mod 10 though?

- Aug 2nd 2010, 10:55 AMundefined
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.

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

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

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

Right, but keep going. What is 7*3 reduced mod 10?

How do you get this? This fails for many y, including y = 2.

What do you think? - Aug 2nd 2010, 11:11 AMsfspitfire23
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). - Aug 2nd 2010, 11:30 AMSoroban
Hello, sfspitfire23!

Quote:

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

Did you crank out a few powers of 56?

. . $\displaystyle \begin{array}{ccc}

56^1 &=& 5\boxed{\!6\!} \\ \\[-4mm]

56^2 &=& 313\boxed{\!6\!} \\ \\[-4mm]

56^3 &=& 175,\!61\boxed{\!6\!} \\ \\[-4mm]

56^4 &=& 9,\!834,\!49\boxed{\!6\!} \\

\vdots && \vdots \end{array}$

Do you see a pattern?

- Aug 2nd 2010, 11:31 AMundefined
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.

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 $\displaystyle 6^2 \equiv 6 \pmod{10}$ 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. - Aug 2nd 2010, 11:50 AMmelese
This may answer your question in general...

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

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

Because $\displaystyle c_i10^i\equiv 0(mod\ 10)$ for $\displaystyle i>0$, we have $\displaystyle N\equiv c_0(mod\ 10)$. This means that if you compute a positive integer modulo 10 and take the least nonnegative residue you find the last digit, $\displaystyle c_0$. - Aug 2nd 2010, 11:51 AMundefined