# Math Help - units/ones digit

1. ## 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?

2. 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?

3. 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?

4. Why mod 10 though?

5. 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.

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

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

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

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

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

Originally Posted by sfspitfire23
Why mod 10 though?
What do you think?

6. 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).

7. Hello, sfspitfire23!

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?

8. 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.

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.

9. 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, $c_0$.

10. 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, $c_0$.