units/ones digit

• Aug 2nd 2010, 10:30 AM
sfspitfire23
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?
• Aug 2nd 2010, 10:41 AM
undefined
Quote:

Originally Posted by sfspitfire23
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?
• Aug 2nd 2010, 10:45 AM
sfspitfire23
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 AM
sfspitfire23
Why mod 10 though?
• Aug 2nd 2010, 10:55 AM
undefined
Quote:

Originally Posted by sfspitfire23
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
We have (7)(3)(y-2) (mod 10)

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

Quote:

Originally Posted by sfspitfire23
So ones place is 1?

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

Quote:

Originally Posted by sfspitfire23
Why mod 10 though?

What do you think?
• Aug 2nd 2010, 11:11 AM
sfspitfire23
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 AM
Soroban
Hello, sfspitfire23!

Quote:

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

Did you crank out a few powers of 56?

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

Do you see a pattern?
• Aug 2nd 2010, 11:31 AM
undefined
Quote:

Originally Posted by sfspitfire23
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
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.
• Aug 2nd 2010, 11:50 AM
melese
Quote:

Originally Posted by sfspitfire23
Why mod 10 though?

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, $c_0$.
• Aug 2nd 2010, 11:51 AM
undefined
Quote:

Originally Posted by melese
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, $c_0$.