Question

What is the general formula for the "tens place" of 3^n where n is greater than or equal to 0? It starts as 0,0,0,2,8,...

Answer 1

Gonna have to think about this one,

Answer 2

Here's a hint: it's periodic. =P

Answer 3

I'm thinking binomial theorem or modular arithmetic. Not like I've studied either of them...

Answer 4

Me either

Answer 5

\(\large \color{black}{ \normalsize \text{Ten's place can be found by}\hspace{.33em}\\~\\
3^{n}\pmod {100} \hspace{.33em}\\~\\~\\~\\
\normalsize \text{or by binomial }\hspace{.33em}\\~\\
3^{n}\hspace{.33em}\\~\\
\implies 9^{n/2}\hspace{.33em}\\~\\
\implies (-1+10)^{n/2}\hspace{.33em}\\~\\
\implies \dbinom{\small{\frac{n}{2}}}{0}(-1)^{n/2}(10)^{0}+\dbinom{\small{\frac{n}{2}}}{1}(-1)^{n/2-1}(10)^{1}\hspace{.33em}\\~\\
}\)

Answer 6

@mathmath333 Should that not be \((3^n \% 100) - (3^n \% 10)\)?

Answer 7

^ and what prevents that from being interpreted as the "general" formula? Huehuehue...

Answer 8

if ur finding \(3^{n}\pmod {100} \hspace{.33em}\\~\\\)

then finding \(3^{n}\pmod {10} \hspace{.33em}\\~\\\)

is just an extra work

\(3^{n}\pmod {100} \hspace{.33em}\\~\\\)

will give u ten's place as well as unit's place

Answer 9

Upm/r1r2=565374.9999

We can all go home now

Answer 10

but you want the tens place only. and for that, the formula would be (3^n % 100 - 3^n % 10)/10

Answer 11

I suppose you can consider \[\Large \frac{3^n \mod 100 - 3^n \mod 10 }{10}\] to be the general formula, but it's not really pretty imo.

Answer 12

if u want only ten's place u can find \(3^{n}\pmod {100} \hspace{.33em}\\~\\\)
and then smartly ignore the last digit ,
no need to waste time on calculating \(3^{n}\pmod {10} \hspace{.33em}\\~\\\)

Answer 13

Haha that's sort of like a function that's the opposite of 1 to 1, it's a purposefully 10 to 0 function =P

Answer 14

0
0
2
8
4
2
8
6
8
4
4
4
2
6
0
2
6
8
6

Answer 15

I think this is on the right path, now we know the period is 20.

Answer 16

Another observation: these numbers correspond to the tens place of \(3^n\) :O
Mindblown.jpg

Answer 17

I actually like the idea of ignoring the last digit... It's kind of a fascinating idea to just throw information away but maybe in a more formal sense. Hmm.

Answer 18

Wait, I thought that was the observation.

Answer 19

What were you thinking when you found these?

Answer 20

How do we generate these? Another exponential function?

Answer 21

I missed one zero in the beginning. Anyway...

Answer 22

@kainui are u trying to solve the ganeshie8's question on triangle perimeter

Answer 23

That's originally how I came across this problem, yes.

Answer 24

But I don't think it is really the best way to solve that problem, and it's its own problem to itself. =P

Answer 25

\(\large \color{black}{\begin{align} \normalsize \text{then this formula might help u}\hspace{.33em}\\~\\

\implies \dbinom{n/2}{0}(-1)^{n/2}(10)^{0}+\dbinom{n/2}{1}(-1)^{n/2-1}(10)^{1}\hspace{.33em}\\~\\
+\dbinom{n/2}{2}(-1)^{n/2-2}(10)^{2}+\dbinom{n/2}{3}(-1)^{n/2-3}(10)^{3}\hspace{.33em}\\~\\

\end{align}}\)

as he asked about the remainder for \(3^n \pmod{10^4}\)

Answer 26

*

Answer 27

Just to mix things up, if \[\Large 3^n = 3^{4a+b}, \ \ \ 0 \le b \le 3\] then\[\Large 10 \lfloor \frac{3^n}{10}\rfloor =8^{b+1}(a-1) \mod 10\] Sadly this is not true for when b=3. But maybe you'll find this interesting to keep you looking for the answer. ;P

Answer 28

latex tip\[a\stackrel{c}{\equiv}b\]:-)