A simple power

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Rushil

1592 posts

Rushil

#1 h Oct 16, 2005, 5:21 AM • 2 Y

Y by Adventure10, Mango247

Prove that the ten's digit of any power of 3 is even.

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tµtµ

393 posts

tµtµ

#2 h Oct 16, 2005, 7:42 AM • 2 Y

Y by Adventure10, Mango247

Ten's = tenth ?

$3^{20} = *3*486784401$

:

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Arne

3660 posts

Arne

#3 h Oct 16, 2005, 8:47 AM • 3 Y

Y by Adventure10, Adventure10, Mango247

No, the tens digit is the last but one digit, the one coming just before the units digit.

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Singular

749 posts

Singular

#4 h Oct 16, 2005, 10:34 AM • 2 Y

Y by Adventure10, Mango247

We are going to look at 3^k mod 20. It repeats 3,9,7,1. Hence result.

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Altheman

6194 posts

Altheman

#5 h Oct 17, 2005, 12:10 AM • 3 Y

Y by tantheta67, Adventure10, Mango247

more completely...

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10000th User

3049 posts

10000th User

#6 h Nov 28, 2005, 10:31 PM • 2 Y

Y by Adventure10, Mango247

Altheman wrote:

$0>n>10$

Small correction: it should be $0<n<10$

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grafitti123

165 posts

grafitti123

#7 h Jul 19, 2009, 2:33 PM • 3 Y

Y by Adventure10, Mango247, DEKT

It can be written in the form:-
$3^{m}=10 \times x+1,3,7,9$
It can easily be shown that if it is 1, then it is 1 mod 4, if it is 3 then it is 3 mod 4, if it is 7 it is 3 mod 4 and if it is 9 then it is 1 mod 4.
(We know that $10x+3=3^1,3^5,3^9....$
$3^{5}\equiv 3\mod{4}$
so,
$3^{9}\equiv 3\mod{4}$
,so on and similarly the others.

Taking each of the four cases , we see that $x\equiv 0\mod{2}$

Yay, my first proper post

.

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soumik

131 posts

soumik

#8 h Jul 19, 2009, 4:51 PM • 2 Y

Y by Adventure10, Mango247

Observe that 3^3=27...
Now after this unit's place of 3^n can be 1,3,9,7,....the cycle repeats, observe each leaves a rem of 0 or 2, which when added to 2 makes it even, thus proved.

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tonotsukasa

12 posts

tonotsukasa

#9 h Apr 26, 2011, 11:16 AM • 1 Y

Y by Adventure10

The last two digit of $3^n$ repeats as
01,03,09,27,81,43,29,87,61,83,49,47,41,23,69,07,21,63,89,67,01,...
Hence $3^n$ 's ten's digits are all even.

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Debdut

528 posts

Debdut

#10 h Feb 13, 2012, 6:45 AM • 3 Y

Y by Ayushakj, Adventure10, and 1 other user

I have a better proof by induction.

Let $N_n = 3^n$
For n=1,
$N_1$ = 03 (ten's digit is even,we will represent even by e)

Assume for n the condition is true that

$N_n$ = _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ e k( k is either of 1,3 ,7,9)

Now, $N_{n+1} = 3*N_n$ = (_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ e k)*3
= 30(_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ e) + 3k
= 10(_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ e) + 3k (since any number multiplied by e is e)
= 10[_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ (e+m)] + k (since 1*3 = 03 , 3*3 = 09 , 7*3 = 21 , 9*3 = 27 so m is always even where m is the ten's digit of 3k)
= _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ e k
[proved]

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madhusudan kale

23 posts

madhusudan kale

#11 h May 12, 2012, 10:04 AM • 2 Y

Y by Adventure10, Mango247

Singular wrote:

We are going to look at 3^k mod 20. It repeats 3,9,7,1. Hence result.

Could you please elaborate.

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Iyerie

163 posts

Iyerie

#12 h Feb 1, 2017, 3:52 AM • 3 Y

Y by bunnyrabbits, Adventure10, Mango247

madhusudan kale wrote:

Singular wrote:

We are going to look at 3^k mod 20. It repeats 3,9,7,1. Hence result.

Could you please elaborate.

I think Singular is referring to the fact that when each of those numbers is multiplied by 3, their tens digit is a 2. Hence the sum of the tens digits must be even is what he is arguing we can infer from that....not entirely sure if that's a rigorous argument.

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Chotu2004

34 posts

Chotu2004

#13 h Aug 13, 2019, 3:33 PM • 2 Y

Y by Adventure10, Mango247

Altheman wrote:

more completely...

what is euler's extension of FLT

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AlastorMoody

2125 posts

AlastorMoody

#14 h Sep 5, 2019, 7:15 PM • 1 Y

Y by Adventure10

RMO 1993 P2 wrote:

Prove that the ten's digit of any power of 3 is even.

Solution: Suppose that the power of $3$ is $3^k$ . We'll do a quick induction on $k$ . First check that $3^3=27$ , $3^4=81$ , $3^5=243$ , $3^6=729$ and $3^7=2187$

Suppose, for some $m \in \mathbb{N}$ , $3^m$ has an even ten's digit, say $x$ . So:
1) If the unit's digit of $3^m$ is 1, then ten's digit of $3^{m+1}$ $\implies$ $3x \equiv \ell \pmod{10}$
2) If the unit's digit of $3^m$ is 3, then ten's digit of $3^{m+1}$ $\implies$ $3x \equiv \ell \pmod{10}$
3) If the unit's digit of $3^m$ is 7, then ten's digit of $3^{m+1}$ $\implies$ $3x+2 \equiv \ell \pmod{10}$
4) If the unit's digit of $3^m$ is 9, then ten's digit of $3^{m+1}$ $\implies$ $3x +2\equiv \ell \pmod{10}$
(Where, $\ell \in \{0,2,4,6,8\}$ ). Hence, our Induction is complete and all powers of $3$ have an even ten's digit! $\qquad \blacksquare$

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leibnitz

1430 posts

leibnitz

#15 h Mar 19, 2020, 12:06 PM

Y by

Really sorry to bump this trivial problem but I thought this approach was nice. $\pmod{20}\Rightarrow$ there exists $t$ such that $20t<3^n<20t+10$ so $2t<\frac{3^n}{10}<2t+1$ hence $[\frac{3^n}{10}]=2t$ ( $[.]$ is the floor function).Note that $[\frac{3^n}{10}]$ is the number formed by removing the last digit of $3^n$ and moreover this is even.The result follows

This post has been edited 3 times. Last edited by leibnitz, Mar 19, 2020, 12:11 PM

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ATGY

2502 posts

ATGY

#16 h Jul 2, 2021, 9:57 AM

Y by

Whoops, I fell asleep while solving this. Here's the solution:

Notice that the last digit of every number in the form $3^k$ (where $k$ is a non-negative integer) goes in the pattern: $3, 9, 7, 1$ .

The last digit will be $3$ when $k \equiv 1\pmod{4}$ , the last digit will be $9$ when $k \equiv 2\pmod{4}$ , the last digit will be $7$ when $k \equiv 3\pmod{4}$ and the last digit will be $1$ when $k \equiv 0\pmod{4}$ . We can separate this problem into 4 different cases.

Case 1: The last digit of $3^k$ is $1$ - This means that $k \equiv 0\pmod{4}$ . $k$ can be written as $4n$ where $n$ is a non-negative integer. So, $3^k = 3^{4n} = 81^n$ . We want the ten's digit to be even, so:
$\frac{81^n - 1}{10} \equiv 0\pmod{2}$ $\implies 81^n \equiv 1\pmod{20}$ This is clearly true as $81 \equiv 1\pmod{20}$ . So we have proved our statement for the first case.

Case 2: The last digit of $3^k$ is $3$ - This means that $k \equiv 1\pmod{4}$ . $k$ can be written as $4n + 1$ and we can proceed to do the same thing as Case 1 to get that $81^n\cdot3^1 \equiv 3\pmod{20}$ which is true.

Case 3: The last digit of $3^k$ is $9$ - This means that $k \equiv 2 \pmod{4}$ . $k$ can be written as $4n + 2$ and we can proceed to do the same thing as Case 1 to get that $81^n\cdot3^2 \equiv 9\pmod{20}$ which is also true.

Case 4: The last digit of $3^k$ is $7$ - This means that $k \equiv 3\pmod{4}$ . $k$ can be written as $4n + 3$ and we can proceed to do the same thing as Case 1 to get that $81^n\cdot3^3 \equiv 7\pmod{20}$ which is true as $81^n \equiv 1\pmod{20}$ and $3^3 \equiv 7\pmod{20}$ .

We are done $\blacksquare$

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SomeonecoolLovesMaths

3198 posts

SomeonecoolLovesMaths

#17 h Sep 16, 2024, 6:25 PM • 2 Y

Y by Nilabha_Sarkar, namanrobin08

We try induction.
Base case: $n=1$ ,
Trivial.
Now let's assume it works for $n=k$ . Proof for $n=k+1$ .
The ten's digit of $3^k$ is even, so multiplying it with $3$ will also result to an even number. So the rest of the role is played by the units digit. Observe the only possible units digits are $1,3,7,9$ . Multiplying them respectively yields $3,9,21,27$ . They too add an even number to the tenth position.As two even numbers add up to an even number, the resultant ten's digit is again even. Hence proved. $\blacksquare$

This post has been edited 1 time. Last edited by SomeonecoolLovesMaths, Sep 16, 2024, 6:25 PM

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