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anonymous
 5 years ago
prove the integer 53^103+103^53 is divisible by 39
anonymous
 5 years ago
prove the integer 53^103+103^53 is divisible by 39

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Are you familiar with modular arithmetic? (I hope).

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0no... :( ...please show me the steps

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.0how do you do this without number theory?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0That's what I'm trying to think about.

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.0If you can tell me the class, maybe I can come up with a way to help you that you will understand

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I'm in 10th...just saw this problem and eager to know how to do it

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.0bond what was your modular arithmetic way?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Well \(53\equiv14\) while \(103\equiv14\), but that's as far as I got.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Plus 53 is an odd power, so the  stays.

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.0oh wait 103 is congruent to 25 and 25+14=39

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.0whenever 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
 5 years ago
Best ResponseYou've already chosen the best response.0Yeah, it's the same idea.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Ok, here's the deal: \[53^{103}\equiv14^{103}=(14^2)^{51}*14\equiv1^{51}*14\equiv14({\rm mod} 39)\]

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.014 is congruent to 1 mod 39?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.014^2=196 and 195 is divisible by 39

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.0jaebond I think I might have to become your fan. I like your number theory skills.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Why thank you, I've been brushing up lately.

myininaya
 5 years ago
Best ResponseYou've already chosen the best response.0Its been awhile since i had number theory

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0So 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
 5 years ago
Best ResponseYou've already chosen the best response.0Most of the properties of the integers hold as normal, except that you usually can't divide.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0So 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
 5 years ago
Best ResponseYou've already chosen the best response.0Does this make sense maria?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0hmmmm.....that's clear enough.... thanks both of you.... :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0why is 103 congruent to 25 ?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Do you know why this is working, Maria?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I would start off by realising 39 is divisible by 3 and 13, then show that 53^103 + 13^53 is divisible by both 3 and 13. It will then be divisible by 39.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0So, 53mod3 == 1 (which means you can write 53 as 3n1, for some integer n). So 53^103 = 3k1, for some integer k. Similarly, 103mod3==1, which means 103^53 can be written 3p+1, for some integer p. When you add them, 3k1+3p+1 = 3(k+p)=3j, for some integer j. Therefore, 3 divides (103^53 + 53^103). That's the first part.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0103mod13==1, so 103^53 = 13s1, 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.
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