How to find the last digit on $3^{100}$? Is there any proper method to solve such questions without calculator of course?
$\endgroup$ 49 Answers
$\begingroup$Solving questions like these are pretty simple once you know a certain trick:
Let's work out the first few terms of $x = 3^n$
When n = 0: | x = 1 | The last digit is 1
When n = 1: | x = 3 | The last digit is 3
When n = 2: | x = 9 | The last digit is 9
When n = 3: | x = 27 | The last digit is 7
When n = 4: | x = 81 | The last digit is 1
When n = 5: | x = 243 | The last digit is 3
There is a pattern! 1,3,9,7,1,3,9,7,1,3,9,7,1,3,9,7,1,3,9,7... and so on. The pattern repeats itself every 4 iterations. {1,3,9,7}{1,3,9,7}...
We can therefore deduce that if:
($n$ mod 4) = 0, The last digit is 1
($n$ mod 4) = 1, The last digit is 3
($n$ mod 4) = 2, The last digit is 9
($n$ mod 4) = 3, The last digit is 7
To work out the final answer of this, let's see what happens when n = 100
$100 $ mod $4 = 0$, Therefore the last digit is a 1! Problem solved!
$\endgroup$ 2 $\begingroup$Hint: $3^4 \equiv 1 \mod 10$.
$\endgroup$ $\begingroup$We have modulo $10$$$3^{4n+k}\equiv\begin{cases}1\text{ if } k=0\\3 \text{ if } k=1\\9 \text{ if } k=2\\7\text{ if } k=3\end{cases}$$ Thus since $100$ is multiple of $4$ $$3^4 \equiv 1 \mod 10$$
$\endgroup$ $\begingroup$Easy way :
$$3^{100}=(3^4)^{25}$$
$$(3^4)^{25}=81^{25}$$
If a number has $1$ as its last digit , any power of such number will have $1$ as its last digit.
So last digit of $81$ is $1$.
$\endgroup$ $\begingroup$If you start exponentiating $3$, you will notice the following alternating pattern for the last digit: $$3,9,7,1,$$ according as the exponent leaves a remainder $1$, $2$, $3$, or $0$, respectively, when divided by $4$. For example, \begin{align*} 3^1=&\,\underline 3,\\ 3^2=&\,\underline 9,\\ 3^3=&\,2\underline 7,\\ 3^4=&\,8\underline 1,\\ 3^5=&\,24\underline 3, \end{align*} and so forth. Since $100\equiv 0(\mod 4),$ the final answer is $1$.
Brute-force answer: $$3^{100}=515\mathord,377\mathord,520\mathord,732\mathord,011\mathord,331\mathord,036\mathord,461\mathord,129\mathord,765\mathord,621\mathord,272\mathord,702\mathord,107\mathord,522\mathord,001.$$
$\endgroup$ 4 $\begingroup$Short & tricky answer, $$10 = (3^2+1) \mid (3^4-1) \mid (3^{100}-1) $$ and since $10$ is a divisor of $3^{100}-1$, the last digit of $3^{100}$ is $\color{red}{1}$.
$\endgroup$ 1 $\begingroup$$3^{100}=(3^4)^{25 }=81^{25}=(80 + 1)^{25}=a*80^{25}+b*80^{24}+.....+x*80^2+y*80 + 1$ where a,b,c,....,y are integer coeffients obtained by expanding $(80 +1)^{25} $. We can calculate these coefficients if we wanted to, but it's enough to know they are integers.
Notice, every term $k*80^j $ is divisible by 10 so
$3^{100}=10M + 1$ where $M= (a*80^{25}+.....+y*80)/10$. $M $ is a whole number, so the last digit is 1.
This is more detail than we need.
It's a basic fact that if we use the notation $a \equiv b \mod n $ to mean that for integers $a,b$ and $n $ that $a$ and $b $ will have the same remainder when divided by $n $ that if $a \equiv 1 \mod n $ then $ak \equiv k \mod n $.
(Because $a \equiv 1 \mod n$ means $a=jn +1$ so $ak =(jk)n+k $ so $ak \equiv k \mod n$.)
Therefore if $a^k \equiv 1 \mod n $ than $a^{mk}=a^k*a^k*....a^k\equiv 1*1*...*1 \mod n $.
So as $3^4=81 \equiv 1 \mod 10$, it follows that $3^{4m} \equiv 1 \mod 10$ and will have 1 as a last digit.
More interesting is what are the last 2 digits of $3^{100} $. $3^4=81 \mod 100$ so $3^8=6400+160+1\equiv 61 \mod 100$. $3^{12} \equiv 81*61 \equiv (6+8)*10+1\equiv 41 \mod 100$. $3^{16}\equiv 61^2\equiv (6+6)10+1\equiv 21\mod 100$. $3^{20}\equiv 21*81\equiv (8+2)10+1 \equiv 1701 \equiv 1\mod 100$. So $3^{20m}\equiv 1 \mod 100$ so the last two digits of $3^{100}$ are 01.
$\endgroup$ $\begingroup$prove that $$3^4\equiv 1 \mod 10$$ then we get $$3^{100}\equiv 1^{25}\equiv 1 \mod 10$$
$\endgroup$ 3 $\begingroup$$$3^{100}=9^{50}=(1-10)^{50}=1-50\cdot10+{50\choose2}10^2-\cdots=1+10\cdot\text{integer}$$
so the last digit is $1$. In fact it's easy to see from this that the last two digits are $01$, and not too hard to see that the last three digits are $001$ (because the first three terms of the expansion are $1-500+62500=62001$).
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