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Math olympiad number theory

  1. May 30, 2017 #1
    • Member warned that some attempt must be shown
    1. The problem statement, all variables and given/known data
    this problem came out in the math olympiad i took today and i got completely wrecked by this

    consider the following equation where m and n are positive integers:

    3m + 3n - 8m - 4n! = 680
    determine the sum all possible values of m:

    2. Relevant equations
    not sure which

    3. The attempt at a solution
    no attempt my mind is completely blank give me some hints please
     
  2. jcsd
  3. May 30, 2017 #2

    scottdave

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    Try moving everything with m to one side and n to the other side. What can you say about the values of the m-function and n function ?
     
  4. May 30, 2017 #3

    BvU

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    Were you allowed to use a calculator ? :smile:
    Code (Text):

        n: 1 2 3 4 5 6
      m 1 -6 -4 -2 -20 -242 -2156
        2 -8 -6 -4 -22 -244 -2158
        3 2 4 6 -12 -234 -2148
        4 48 50 52 34 -188 -2102
        5 202 204 206 188 -34 -1948
        6 680 682 684 666 444 -1470
     
  5. May 30, 2017 #4
  6. May 30, 2017 #5

    BvU

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    Just kidding. But from the brute force results you should be able to extract some restrictions on the (m,n) pairs to check. Especially in combination with a list of powers of 3:
    3
    9
    27
    81
    243
    729

    and a list of 4n! helps a lot too!
    4
    8
    24
    96
    480
    2880
    These don't take long to set up, even without a calculator.

    subsequent hypotheses:
    • one of the two has to be 6
    • m has to be 6, n has to be smaller than 5

    I think the hard part is to find the one (6,1) combination that does yield 680. Then 'conclude' that there can be no others.
     
  7. May 30, 2017 #6
    @BvU
    not to offend you but i am not really fan of brute force approaches

    and why do you say n has to be 6

    based on what scottdave said

    i inferred m > n is this true
    and if it is does it help to write m = k + n
     
  8. May 30, 2017 #7
    ##3^m + 3^n = 680 + 8m + 4n!## or ##3^m + 3^n = 4(170 + 2m + n!)##

    Since ##170 + 2m + n!## is divisible by ##3## what does this tell you about parity of ##170 + 2m + n!## ? and for what values of ##n## it is possible ?
     
  9. May 30, 2017 #8

    BvU

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    No offence taken. I'm just interested, and to be sure: I don't know the proper way to deal with this one, so I just did trial and error. 4n! goes real fast as you can see, so n can't be 6. You need at least one of the two with a value of 6 to get over 680 .

    I hadn't pulled out the factor 4 Buffu mentions. Hats off !
     
  10. May 30, 2017 #9
    oh wow that is interesting so 2m + n! ≡ 1 mod3

    if n≥3 then m must be multiple of 3

    if n <3 then 2m ≡ 2 mod 3

    am i getting somewhere
     
  11. May 30, 2017 #10
    @BvU
    and more over i am a really careless boy i am sure to make lots of mistakes i probably lost 8 questions just by careless today so that's another reason i did not opt for your method without a calculator so ya anyways trial and error to estimate bounds and restrictions is always uselful
     
  12. May 30, 2017 #11
    You can do what you are doing but an easy way would be to check if ##170 + 2m + n!## is odd or even ?
     
  13. May 30, 2017 #12
    oh is n't that always even for n bigger than 2

    but what does that tell you the left hand side is also always even
     
  14. May 30, 2017 #13
    But we want it odd, so what values of ##n## is available to us ?
     
  15. May 30, 2017 #14
    why do we want it odd? please i am beginner in olympiads

    but anyways the answer to your question is n =1

    edit :

    oh sorry now i get why it must be odd because it is divisible by 3 sorry for being so dumb

    so n= 1 to give 170 + 8m + 1! to be odd

    3m = 680 + 8m + 1

    so m must be at least 6 and so is it only m = 6 solves it

    that is very brilliant method there how did you get to it did you just play around with the expression until you got something useful
     
    Last edited: May 30, 2017
  16. May 30, 2017 #15
    Nooo, It is wrong, since even multiples of 3 are there :).

    I said that expression is odd because ##3^n + 3^m## is not divisible by 8 for any value of ##m,n## * and if ##170 + 8m + n!## is even, then ##3^m + 3^n## must be divisible by ##8##.

    * ##3^n \mod 8 = 1, 3## using power rule for modulus and the fact that ##3^n = 1 ## for ##n = 0##.

    So ##3^n + 3^m \mod 8 = 2, 4, 6## hence never divisible by 8.
     
  17. May 30, 2017 #16
    I remember some tricks from olympiads when I used to give those.
     
  18. May 31, 2017 #17
    oh okay now i get it thanks for the help
     
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