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Remainders by modular arithmatic

  1. Feb 23, 2013 #1
    find the remainders for

    a) 2293[itex]\equiv[/itex] x (mod 15)

    b) 243101[itex]\equiv[/itex] x(mod 8)

    c) 52001 + (27)! [itex]\equiv[/itex] x (mod 8)


    a) I was able to equate 2[itex]\equiv[/itex]-13(mod15) ==> 22[itex]\equiv[/itex]4(mod15)

    my idea here was to get 29 and then multiply by a power of 100 which would give me 900 and then work out the other 23 from there, but I've already hit a road block....I'm going to end up with a huge number just from the product of the moduluses......maybe that's what is suppose to happen, but I'm under the assumption that I'm suppose to figure these out without a calculator.

    b) I was able to work it down to a congruence of 243 [itex]\equiv[/itex] 3(mod 8) but if I raise that to the 101 on both sides.....again I'm out of luck in terms of the 3.

    help please.......
     
  2. jcsd
  3. Feb 23, 2013 #2

    Dick

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    Try going up in larger steps by using powers. 2^2=4 mod 15. Square both sides, so 2^4=4^2=1 mod 15. You can get up to 2^8, 2^16 pretty easily by repeated squaring and taking mods. But look at 2^4=1 mod 15. That's pretty useful!
     
  4. Feb 23, 2013 #3
    Thanks I actually figured it out with that hint. I had one more question, but I didn't want to start a new thread for it:

    Show that their is no digit "a" such that the number 2794a1 is divisible by 8.

    So I tried to apply the rule that "every natural number is congruent to the sum of its digits modulo 9"

    Now I know this isn't 9, but I tried this:

    2 x 105 + 7 x 104 +.....+ a x 10 + 1

    working out the mods of the 10's I got

    10 [itex]\equiv[/itex] 2(mod 8)
    102[itex]\equiv[/itex] 4(mod 8)
    103, 104, 105 were all [itex]\equiv[/itex] 0 (mod 8)

    I have a feeling the other two are as well but I don't think I want to deal with negatives.

    Doing this and combining the info I obtained this expression after multiplying by the modulus and following the rules:

    = 4+a+1
    = 5 + a (mod 8)

    but obviously there are some values that will allow this to be divisble by 8, so I messed up somewhere.
     
  5. Feb 23, 2013 #4

    Dick

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    Yeah, you messed up. You don't just want to add the digits, do you? The value mod 8 will be the value of each of the digits times the mod value of the corresponding power of 10. What equation do you really want to solve?
     
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