Question

Math quest(For fun)

Answer 1

What is the digit in the unit place of \(2003^{2001}\)?

Answer 2

Can I use a calculator? xD

Answer 3

u can't use calculator X'D
U will get infinity X'D

Answer 4

I have a super-calculator ;)
Anyways I'll try it without it. I'm starting with \[2003^{2000}2003^1\]

Answer 5

I also notice that 3^n has a unit digit pattern of 3 9 7 1 3 9 7 1 3 9 7 1...

Basically the pattern is 4 repeating numbers.

so 2000 mod 4 is 1, and the 1st number of that pattern is 3. That multiplied by 3 (for the 2003^1) is 9.

So my guess will be 9. Am I correct?

Answer 6

what about the second digit?

Answer 7

fyi 2000 mod 4 is 0

Answer 8

the first 4 digits are 2003

Answer 9

9 is incorrect but quiet close to the real answer,hehe.
Good guess tho,mhchen.

nuts,I don't get what u mean. X'D

yeah, Zarkon was right.
2000 mod 4 is 0

To find the unit digit you only need to look at the unit digit and the power and not the whole number.
So, unit digit of \(2003^{2001}\) will be same as unit digit of \(3^{2001}\).

Now if we look at the unit digit of powers of any number they follow a cycle i.e. They repeat itself after a fixed interval.

Let us look at powers of 3.

\(3^0=1\)

\(3^1=3\)

\(3^2=9\)

\(3^3=27\)(unit digit is \(7\))

\(3^4 = 81\) (unit digit is \(1\), repetition starts)

\(3^5= 243\)(unit digit is \(3\), again repetition)

Likewise we can go on..
The observation is that the unit digit repeats itself at an interval of \(4\), hence if we divide the power by four and check the unit digit for \(3\) raised to the power remainder obtained.. we will get our answer.

Hence, \(\frac{2001}{4}\) gives remainder \(1\)

Unit digit of \(2003^{2001}\) should be same as unit digit of \(3^2001\) which would be same as \(3^1\) which is 3.

Hence the answer is 3.

Answer 10

you can use a generalization of Fermat's little theorem to get the answer I gave for the first 4 digits ... which is 2003

\[2003^{2001}\equiv 2003 \,( \hspace{-.4cm}\mod10000)\]

which obviously gives the ones digit. 200\(\color{red}{3}\)

Answer 11

Wow,that's nice,Zarkon. :D
Learn something new today,hehe,thanks. ;)

Compute some powers of 2003 mod 10:
\(2003^{2}\equiv3^{2}\equiv9(mod~10)\)
so,
\(2003^{4}\equiv9^{2}\equiv1(mod~10)\)

Therefore,\(2003^{2001}\equiv(2003^4)^{500}\times2003\) and
\(2003^{2001}\equiv(1)^{500}\times2003(mod~10)\)
\(\equiv1\times2003(mod~10)\)
\(\equiv2003(mod~10)\)
\(\equiv3(mod~10)\)

Zarkon,is my working correct or?

Answer 12

that is correct

Answer 13

Cool!
Thank you,Zarkon! :D

lol,thanks guys...

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