Math quest(For fun)

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

7 months agoCan I use a calculator? xD

7 months agou can't use calculator X'D U will get infinity X'D

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

7 months agoI 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?

7 months agowhat about the second digit?

7 months agofyi 2000 mod 4 is 0

7 months agothe first 4 digits are 2003

7 months ago9 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.

7 months agoyou 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\(\color{red}{3}\)

7 months agoWow,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?

7 months agothat is correct

7 months agoCool! Thank you,Zarkon! :D lol,thanks guys...

7 months ago