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Math Help - R(n)=4n+3 Complicated Word Problem

  1. #1
    Senior Member Mukilab's Avatar
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    R(n)=4n+3 Complicated Word Problem

    Hello, I'm having trouble with a question frmo the British mathematical Olympiad (1997)

    N is a four-digit integer which does not end in a zero, and R(n) is the four digit integer obtained by reversing the digits of N. E.g. R(3275)=5723

    Determine all such integers N for which R(N)=4N+3

    I used the year (1997) as a reference and I found out it does work for the above statement so that's one but I don't know how many others there are.

    I've worked out that according to the above statement, R(N) would have to start with either 1,3,5,7 or 9

    I've tried laying it out so N=1000a +100b+10c+d and because R(N) must be odd due to +3 it would =4000d+400c+40b+4a and therefore a must be odd, narrowing down the first and last integer of both N and R(N) to be odd.


    Now I am stuck. Please help :/
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  2. #2
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    Quote Originally Posted by Mukilab View Post
    I've tried laying it out so N=1000a +100b+10c+d and because R(N) must be odd due to +3 it would =4000d+400c+40b+4a ....
    That'll simplify to d = (332a + 20b - 130c -1) / 1333

    1997 : 7991 is ONLY integer solution.
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  3. #3
    Senior Member Mukilab's Avatar
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    Explanation?

    I've gotten that sum... it doesn't prove anything...
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  4. #4
    MHF Contributor
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    Quote Originally Posted by Mukilab View Post
    Explanation?
    I've gotten that sum... it doesn't prove anything...
    Not every "puzzle" has a proof...
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  5. #5
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    Since d is an integer (between 0 and 9), 332a+ 20b- 130c- 1
    must be one of

    1: 0
    2: 1333
    3: 2(1333)= 2666
    4: 3(1333)= 3999
    5: 4(1333)= 5332
    6: 5(1333)= 6665
    7: 6(1333)= 7998
    8: 7(1333)= 9331
    9: 8(1333)= 10664
    10: 9(1333)= 11997

    That should help.
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