MARC:

Math quest(For fun)

2 months ago
MARC:

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

2 months ago
mhchen:

Can I use a calculator? xD

2 months ago
MARC:

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

2 months ago
mhchen:

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

2 months ago
mhchen:

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?

2 months ago
nuts:

what about the second digit?

2 months ago
Zarkon:

fyi 2000 mod 4 is 0

2 months ago
Zarkon:

the first 4 digits are 2003

2 months ago
MARC:

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.

2 months ago
Zarkon:

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\(\color{red}{3}\)

2 months ago
MARC:

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?

2 months ago
Zarkon:

that is correct

2 months ago
MARC:

Cool! Thank you,Zarkon! :D lol,thanks guys...

2 months ago