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?
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?
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?
Hello, sfspitfire23!
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?
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.
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, $\displaystyle c_0$.