MARC:

Math quest(For fun)

3 weeks ago
MARC:

What is the digit in the unit place of $$2003^{2001}$$?

3 weeks ago
mhchen:

Can I use a calculator? xD

3 weeks ago
MARC:

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

3 weeks ago
mhchen:

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

3 weeks 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?

3 weeks ago
nuts:

what about the second digit?

3 weeks ago
Zarkon:

fyi 2000 mod 4 is 0

3 weeks ago
Zarkon:

the first 4 digits are 2003

3 weeks 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.

3 weeks 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$$\color{red}{3}$$

3 weeks 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?

3 weeks ago
Zarkon:

that is correct

3 weeks ago
MARC:

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

3 weeks ago