anonymous
  • anonymous
A not so easy problem: Find the remainder when \( 2^{400} \) is divided by \( 400 \)?
Mathematics
katieb
  • katieb
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LollyLau
  • LollyLau
theres a rule for finding remainders... lemme check.
anonymous
  • anonymous
Wow, is that so? Can you please enlighten me? :)
LollyLau
  • LollyLau
176. (may be wrong)

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LollyLau
  • LollyLau
the number cycles every 20 numbers. i tried to simplify the exponent but sadly that didn't work :p
asnaseer
  • asnaseer
I don't know if this counts as a proof but here goes. We can first simplify the problem by dividing numerator and denominator by 4 to get:\[\frac{2^{400}}{400}=\frac{2^{398}}{100}\]This implies we just need to find the last two digits of \(2^{398}\) and then multiply that by 4 to get the desired answer. Now, listing the last two digits of successive multiples of two (starting from 2), we notice the following pattern: 02, 04, 08, 16, 32, 64, 28, 56, 12, 24, 48, 96, 92, 84, 68, 36, 72, 44, 88, 76, 52, 04, 16, ^ pattern starts repeating from 04 (i.e. initial 02 is never encountered again) So, after skipping the initial 02, we therefore have a repeating pattern after every 20 digit pairs. I therefore used this to calculate the position of the last two digits:\[1 + (398-1)\%20=1+17=18\]and the 18th digit pair in the sequence above is 44. Therefore, the remainder must be \(4*44=176\) which agrees with @LollyLau.
anonymous
  • anonymous
You can't divide the numbers like that :)
asnaseer
  • asnaseer
?
anonymous
  • anonymous
"We can first simplify the problem by dividing numerator and denominator by 4 to get" This won't always for remainders.
anonymous
  • anonymous
work*
asnaseer
  • asnaseer
surely it must work? if, after re-multiplying the new remainder you get a number larger than 400, then you would just need to mod it with 400 and that should still work shouldn't it?
asnaseer
  • asnaseer
maybe I don't understand number theory well enough to understand that?
anonymous
  • anonymous
Yes if you again do the mod then it should work.
asnaseer
  • asnaseer
So is my solution valid and correct?
anonymous
  • anonymous
But there are some pretty fine lines here, and I am not sure if this in in general.
anonymous
  • anonymous
What I know, we can't cancel unless the factors are coprime.
asnaseer
  • asnaseer
Sounds like time for another lesson for me. Do you know any good online resources for this topic in particular?
anonymous
  • anonymous
Hm, sorry I don't know any for this in particular :(
asnaseer
  • asnaseer
Ok - no problems - time to google then... :-)
asnaseer
  • asnaseer
BTW: The digit pair I wrote in the list above should be 08 and NOT 16
anonymous
  • anonymous
Wait, I am pretty sure that you can't cancel like that and then multiply, what you can do is break up into coprime factors and then do something like this, why co-prime works should be evident from CRT.
asnaseer
  • asnaseer
*The last digit pair
anonymous
  • anonymous
Btw this problem is good example for using http://en.wikipedia.org/wiki/Chinese_remainder_theorem
anonymous
  • anonymous
Break 400 as 25 and 16 and then apply CRT, note (25,16)=1; so it works :)
asnaseer
  • asnaseer
OK - Let me go learn some chinese then... :-) Thanks again
anonymous
  • anonymous
It's an awesome tool :) Good luck ansaseer :)

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