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anonymous

  • 5 years ago

prove the integer 53^103+103^53 is divisible by 39

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  1. anonymous
    • 5 years ago
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    Are you familiar with modular arithmetic? (I hope).

  2. anonymous
    • 5 years ago
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    no... :( ...please show me the steps

  3. anonymous
    • 5 years ago
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    Oh.

  4. myininaya
    • 5 years ago
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    how do you do this without number theory?

  5. anonymous
    • 5 years ago
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    That's what I'm trying to think about.

  6. myininaya
    • 5 years ago
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    what class is this?

  7. myininaya
    • 5 years ago
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    If you can tell me the class, maybe I can come up with a way to help you that you will understand

  8. anonymous
    • 5 years ago
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    I'm in 10th...just saw this problem and eager to know how to do it

  9. myininaya
    • 5 years ago
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    bond what was your modular arithmetic way?

  10. anonymous
    • 5 years ago
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    Well \(53\equiv14\) while \(103\equiv-14\), but that's as far as I got.

  11. anonymous
    • 5 years ago
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    Plus 53 is an odd power, so the - stays.

  12. myininaya
    • 5 years ago
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    oh wait 103 is congruent to 25 and 25+14=39

  13. myininaya
    • 5 years ago
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    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

  14. anonymous
    • 5 years ago
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    Yeah, it's the same idea.

  15. anonymous
    • 5 years ago
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    \(-14\equiv25\)

  16. myininaya
    • 5 years ago
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    your right though

  17. anonymous
    • 5 years ago
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    Ok, here's the deal: \[53^{103}\equiv14^{103}=(14^2)^{51}*14\equiv1^{51}*14\equiv14({\rm mod} 39)\]

  18. myininaya
    • 5 years ago
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    14 is congruent to 1 mod 39?

  19. anonymous
    • 5 years ago
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    no, but 14^2 is

  20. anonymous
    • 5 years ago
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    14^2=196 and 195 is divisible by 39

  21. myininaya
    • 5 years ago
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    jaebond I think I might have to become your fan. I like your number theory skills.

  22. anonymous
    • 5 years ago
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    Why thank you, I've been brushing up lately.

  23. myininaya
    • 5 years ago
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    Its been awhile since i had number theory

  24. anonymous
    • 5 years ago
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    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 \(14\equiv53\equiv92 ({\rm mod }39)\)

  25. anonymous
    • 5 years ago
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    Most of the properties of the integers hold as normal, except that you usually can't divide.

  26. anonymous
    • 5 years ago
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    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 \(14*14*14*...*14=14^{103}\).

  27. anonymous
    • 5 years ago
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    Does this make sense maria?

  28. anonymous
    • 5 years ago
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    hmmmm.....that's clear enough.... thanks both of you.... :)

  29. anonymous
    • 5 years ago
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    why is 103 congruent to 25 ?

  30. anonymous
    • 5 years ago
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    Do you know why this is working, Maria?

  31. anonymous
    • 5 years ago
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    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.

  32. anonymous
    • 5 years ago
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    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.

  33. anonymous
    • 5 years ago
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    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.

  34. anonymous
    • 5 years ago
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    ohhhhhhhh.....I see

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