anonymous
  • anonymous
prove the integer 53^103+103^53 is divisible by 39
Mathematics
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
Are you familiar with modular arithmetic? (I hope).
anonymous
  • anonymous
no... :( ...please show me the steps
anonymous
  • anonymous
Oh.

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myininaya
  • myininaya
how do you do this without number theory?
anonymous
  • anonymous
That's what I'm trying to think about.
myininaya
  • myininaya
what class is this?
myininaya
  • myininaya
If you can tell me the class, maybe I can come up with a way to help you that you will understand
anonymous
  • anonymous
I'm in 10th...just saw this problem and eager to know how to do it
myininaya
  • myininaya
bond what was your modular arithmetic way?
anonymous
  • anonymous
Well \(53\equiv14\) while \(103\equiv-14\), but that's as far as I got.
anonymous
  • anonymous
Plus 53 is an odd power, so the - stays.
myininaya
  • myininaya
oh wait 103 is congruent to 25 and 25+14=39
myininaya
  • myininaya
whenever i took number theory we wrote the numbers could only be between 0 and whatever number minus 1 we are writing are congruences in
anonymous
  • anonymous
Yeah, it's the same idea.
anonymous
  • anonymous
\(-14\equiv25\)
myininaya
  • myininaya
your right though
anonymous
  • anonymous
Ok, here's the deal: \[53^{103}\equiv14^{103}=(14^2)^{51}*14\equiv1^{51}*14\equiv14({\rm mod} 39)\]
myininaya
  • myininaya
14 is congruent to 1 mod 39?
anonymous
  • anonymous
no, but 14^2 is
anonymous
  • anonymous
14^2=196 and 195 is divisible by 39
myininaya
  • myininaya
jaebond I think I might have to become your fan. I like your number theory skills.
anonymous
  • anonymous
Why thank you, I've been brushing up lately.
myininaya
  • myininaya
Its been awhile since i had number theory
anonymous
  • anonymous
So maria, the idea behind modular arithmetic is that we say that two numbers are the same if they differ by a multiple of modulus. In this case, if the "modulus" is 39, we say that 14 and 53 are the same. We also say that 92 is the same and we can write \(14\equiv53\equiv92 ({\rm mod }39)\)
anonymous
  • anonymous
Most of the properties of the integers hold as normal, except that you usually can't divide.
anonymous
  • anonymous
So the first step from what I did earlier looks like \[53^{103}=53*53*53*...*53\] where there are 103 53's. But \(53\equiv14\), so this becomes \(14*14*14*...*14=14^{103}\).
anonymous
  • anonymous
Does this make sense maria?
anonymous
  • anonymous
hmmmm.....that's clear enough.... thanks both of you.... :)
anonymous
  • anonymous
why is 103 congruent to 25 ?
anonymous
  • anonymous
Do you know why this is working, Maria?
anonymous
  • anonymous
I would start off by realising 39 is divisible by 3 and 13, then show that 53^103 + 1-3^53 is divisible by both 3 and 13. It will then be divisible by 39.
anonymous
  • anonymous
So, 53mod3 == -1 (which means you can write 53 as 3n-1, for some integer n). So 53^103 = 3k-1, for some integer k. Similarly, 103mod3==1, which means 103^53 can be written 3p+1, for some integer p. When you add them, 3k-1+3p+1 = 3(k+p)=3j, for some integer j. Therefore, 3 divides (103^53 + 53^103). That's the first part.
anonymous
  • anonymous
103mod13==-1, so 103^53 = 13s-1, for some integer s. 53mod13==1, so 53^13= 13t+1, for some integer t. Adding them, 103^53+53^13=13(s+t)=13r, for some integer r. Therefore, 13 divides your sum. Since your sum is divisible by 3 and 13, and 3 and 13 are prime, it's divisible by 39.
anonymous
  • anonymous
ohhhhhhhh.....I see

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