20 of 100: So Many Threes

Since $3^4=8\color{#D61F06}1$ , hence $3^4$ raised to any integral power would end in $\color{#D61F06} 1$ only since $1$ raised to any power would be $1$ only or $(3^4)^n$ has it's unit digit as $\boxed{\color{#D61F06}1}$ . For this question $n=2500$ .

This pretty much explains the cyclicity in the unit digits when $3$ is raised to any power. For $n\in Z^{+}\cup \{0\}$ , unit digit of the number of the form: $3^{4n+1}=3^{4n}\cdot 3^1=\underbrace{3^{4n}}_{\text{ends in }\color{#D61F06}1}\cdot 3 \text{ will be } \color{#3D99F6} 3$ $3^{4n+2}=3^{4n}\cdot 3^2=\underbrace{3^{4n}}_{\text{ends in }\color{#D61F06}1}\cdot 9 \text{ will be } \color{#3D99F6} 9$ $3^{4n+3}=3^{4n}\cdot 3^3=\underbrace{3^{4n}}_{\text{ends in }\color{#D61F06}1}\cdot 27 \text{ will be } \color{#3D99F6} 7$

Bonus Challenge(Generalisation):-

Note $\underbrace{999\cdots 99}_{\color{#20A900}\text{n times}}=10^n-1$ $\begin{aligned}\therefore \underbrace{999\cdots 99}_{\color{#20A900}\text{n times}}\times 9=&(10^n-1)\times 9\\=&10^n\times 9-9\\=&9\underbrace{000\cdots 00}_{\color{#20A900}\text{n times}} -9\\=& 8\underbrace{999\cdots 99}_{\color{#69047E}\text{(n-1) times}} \color{#333333}1\\&(\text{Digitwise subtraction})\end{aligned}$

So for $n=10000$ , $\underbrace{9999 \cdots 9}_{\text{10000 9's}}\ \times 9$ will contain $\color{#69047E}10000-1=9999$ $\text{9's}.$

We can see that

3^1 = 3

3^2 = 9

3^3 =27

3^4=81

3^5 =243

Therefore the pattern is that

When power is of 4k+1 form last digit is 3

When power is of 4k+2 form last digit is 9

When power is of 4k+3 form last digit is 7

When power is of 4k form last digit is 1

Here power is 10000 which is of 4k type

So last digit is 1

$3^2 = -1 \mod 10$ $\implies 3^{10000} = 1 \mod 10$ [Ans]

For the bonus question: 9x9=81 99x9=891 999x9=8991 Following this pattern, there is always 1 less 9 in the product than in the original number. So, if we have 10,000 digits of 9, the final answer will have 9999 9's.

Let $\phi(x)$ denote the Euler Totient Function. Note that $\phi(10)=4$ . Thus, if $\gcd(a,10)=1$ , then by Euler's Theorem, we have

$a^4 \equiv 1 \pmod{10}$

We find that this is true for $a=3$ . Now, onto the question. Note that

$\begin{aligned} 3^{10000} &\equiv (3^4)^{2500} \pmod{10}\\ &\equiv 1^{2500} \pmod{10}\\ &\equiv 1 \pmod{10} \end{aligned}$

Thus, the final digit is 1.

This reminded me of powers of i!

For this problem, I wrote out powers of 3 until 3^8. (I ran out of space on my desk... whoops!) I found that the last digit is the same for every fourth power. I Because 10000/4 has a remainder of 0, I reasoned that--like 3^0, 3^4 and 3^8-the last digit must be 1.

I'm psyched to have gotten a number theory question right! I'm usually fairly iffy about it!

For the bonus question, there's a beautiful theory I want to share. Usually we do multiply from the last digit, you can't tell how a specific digit presents until a lot of calculations are done. This question can be a example.

We first assume that the result has one digit more.(it can be 0 if necessary) Then put a decimal point after the first result digit. (0.999...x9=?.nnnn) Because 0.999...is between 8/9 and 9/9, the first digit should be 8, because 9 would be too big, 7 would be too small. Repeat the process(individually, different from the traditional process), the second digit and the follow are a bit different, there's already stands a digit to be multipled. It deserves a 9x9=...1,the second digit is 8(from earlier process)+1=9... the last digit has no come up, is 1+0=1. Problem completes now, 9999.

Examples about this theory: 071589x3 First digit:0.71589 is between 2/3 and 3/3....=2; Second digit:7x3 deserves 1, 0.1589 is between 0/3 and 1/3....1+0=1 Third:(1x3deserves)3+(between1/3and2/3)1=4; Fourth:(deserve)5+(between)2=7; Fifth:4+2=6; Sixth=7+0=7; Answer:214767

More examples: 06465x8 0+5=5;( 5/8=0.625,6/8=0.75) 8+3=1;(6x8=8;3/8=0.375,4/8=0.5) 2+5=7;(4x8=2;5/8=0.625,6/8=0.75) 8+4=2; 0+0=0; Answer:51720

012996x5 Comparing fractions:1/5=0.2,2/5=0.4... 0+0=0; 5+1=6; 0+4=4; 5+4=9; 5+3=8; 0+0=0 Answer:064980

Another version: Comparing fraction: 1/2=0.5 2x 087769872 Deserve: 064428644 Between:111111100 Sum: 175539744

Comparing fractions: 1/7=0.142857...2/7=0.285714... 3/7=0.428571...4/7=0.571428... 5/7=0.714285...6/7=0.857142... 7x 0299895794932 Deserve: 0433635938314 Between:2666645636210 Sum: 2099270564524

As you see, this method has a decisive advantage when the number gets big. (...in saving paperwork?)

3 power 4 ends with 1, so if we can divide 1000 by 4 without a reminder we get 1 on the end of result number. It's simple. and bonus chalange: we begin with small numbers, 9 * 9=81, 99 * 9=891, 999 * 9=8991, so we get induction. The result is 9999 9's.

pattern of last terms is 1(for 3^0), 3(for 3^1), 9(for 3^2), 7(for 3^3) which means that you have to divide the n exponent by 4 and the remainder will be either 0, 1, 2 or 3, which would tell you which is the last term of the operation...

3,9,7,1 10000/4=2500(r0) is1

• 3^1 = 3,
• 3^2 = 9,
• 3^3 = 2 7 ,
• 3^4 = (just multiply the ones digit from before by 3) . 1
• 3^5 = (just multiply the ones digit from by 3) .. 3

In this pattern, it is found that every ones digit starts from 3, 9, 7, 1, and repeats on and on. If you want to be sure about this pattern, just multiply the ones digit by 3 from the number before, and you will notice these 4 numbers constantly repeating each other.
So, 1000(the power)/4(the cycle of numbers till a repeat) = 2500. Since the answer is a whole number, 1 , the last number in the pattern of the four numbers, is the answer. So one is the answer

This was the solution I gave for a similar question on quora !! Hope u like it It's the general method of solving any type of question ...involving cyclicity Here's the link for the complete answer .. https://www.quora.com/What-is-the-units-place-of-2013-2016/answer/P-Ramyashri?share=cbea2b6c&srid=h53Or

3^10000 = (3^2)^5000

3^2 = 9 = -1 (mod 10)

Bonus challenge: 9999 digits

Cyclicity As 10000 when divided by 4 leave remainder 0 it's last got will be 4

This is a one liner.(If you are familiar to modular arithmetic). Finding the last digit means to find the remainder when the given number is divided by 10, or in other words find the value of amod10 where a is the given no.

Here we need to find $3^{10000}$ mod10.

Since 3 and 10 are co prime,

$3^{\phi(10)}\equiv1\mod10$ [As a result of Euler's Theorem]

or, $3^4\equiv1\mod10$

or, $3^{10000}\equiv1\mod10$ [Raising both sides to the 2500th power]

3^1=3, 3^2=9,3^3=27,3^4=81,3^5=243. Notice the cycle is 4 long and the last possible digits are 3,9,7, and 1. You can easily see 5 is wrong. Since 10,000 is obviously divisible by 4, it will have the same units digit as 3^4, which is 1.

note that the last digit of 3**n cycles through the the sequence (1,3,9,7) and that 1000 mod 4 = 0 and the 0th value of the sequence is 1

For 9**1000 we get a large number which contains 100 repetitions of the digit '9'

3^10000 = 9^5000

The last digit of 9 raised to any number will have only two possible numbers .

They are 1 & 9.

When the power is odd, (like 9^1,9^3... etc ) then the last digit is 9.

When the power is even, (like 9^2,9^4,9^6... etc) then the last digit is 1.

So the last digit of 3^10000 is 1

We know that $3^4 = 81$

And 10000 is divisible by 4 (i.e 4*2500 = 10000), therefore:

$3^{ 4 }\quad \equiv \quad 1\quad (mod\quad 10)\\ \Rightarrow \quad 3^{10000} \equiv 1 \quad (mod \quad 10)$

The last digit of $3^{10000}$ is $\boxed{1}$

The problem is equivalent to finding $3^{10000} \bmod 10$ .

$\begin{aligned} 3^{10000} \equiv 3^{2\times 5000} \equiv 9^{5000} \equiv (10-1)^{5000} \equiv (-1)^{5000} \equiv \boxed{1} \text{ (mod 10)} \end{aligned}$

The last digits of the powers of 3 are cyclical with a period of 4: 3^0=1, 3^1=3, 3^2=9, 3^3=x7. Since 3^(4k)=xxx1 and 10000 is a multiple of 4, 3^10000=xxxxxxxxx1.

Bonus question - 9999.

Just see that $\left(3^5 \right)^{2000} \equiv 1^{2000} \equiv 1 \pmod {10}$ . That's all.

For the bonus challenge take $\left(10^{10000}-1 \right) \cdot 9 = 9 \cdot 10^{10000}-9 = 899 \dots 1$ . So the digit $9$ appears $9999$ times.

$\boxed{1}$ You can use Euler's theorem ,(Euler totient function) like Sharky has very good done.

$\boxed{2}$ Another way, $3^2= 9 \equiv (-1) \text{ mod(10)} \Rightarrow 3^{10000} = (3^2)^{5000} \equiv (-1)^{5000} = 1 \text{ mod(10)}$

$\boxed{3}$ Use Charmichael function $3^4 =81 \equiv 1 \text{ mod (10)} \Rightarrow 3^{10000}= (3^4)^{2500} \equiv 1^{2500} = 1 \text{ mod(10)}$

$\boxed{4}$ Use Bezout identityand Euler algorithm and Newton Bynomial theorem... $10 - 3 \cdot 3 = 1 \Rightarrow ·3^2 = 10 - 1 \Rightarrow 3^{100000} = (3^2)^{50000} = (10 - 1)^{50000} = {50000 \choose 0} 10^{50000} - {50000 \choose 1} 10^{49999} + \cdots + {50000 \choose 50000} 10^{0} \equiv 1 \text{ mod(10)}$

$\boxed{5}$ Use chinese remainder theorem (exercise for the reader) $3 \equiv 1 \text{ mod(2)}$ , $3 \equiv -2 \text{ mod(5)}...$

But why is it cyclic. and why 4?

It follows the powers of i and the subsequent decimals generated by the power when divided by 4. i^1=3^1=1/4=.25 Last digit is 3 i^2=3^2=2/4=.50 Last digit is 9 i^3=3^3=3/4=.75 Last digit is 7 i^4=3^4=4/4=1.0 Last digit is 1

Then it started all over again. So i^1000=3^1000=1000/4=250.0 So the last digit is 1. In summary, just divide the power by 4 and see the corresponding demicals as shown above.