Golden Year Puzzle

Standard

This is a puzzle proposed by D. R. Kaprekar in year1984.

In the year 1949, a man turned 67. His four sons turned 37, 31, 29, 23. All the five reached prime age in a prime year. It was the golden year for the family. When was or will be their next golden year ?

(Prob 8-3-1, Function, Col 8, Dept of Mathematics, Monash University, Claighton, Victoria, 3168, July 84, pg. 28)

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3 responses »

  1. Hereby I am giving a method to get solutions for this problem.

    Note that

    23\equiv 1 \bmod{2}
    23\equiv 2 \bmod{3}
    23\equiv 3 \bmod{5}
    23\equiv 2 \bmod{7}

    29\equiv 1 \bmod{2}
    29\equiv 2 \bmod{3}
    29\equiv 4 \bmod{5}
    29\equiv 1 \bmod{7}

    31\equiv 1 \bmod{2}
    31\equiv 1 \bmod{3}
    31\equiv 1 \bmod{5}
    31\equiv 3 \bmod{7}

    37\equiv 1 \bmod{2}
    37\equiv 1 \bmod{3}
    37\equiv 2 \bmod{5}
    37\equiv 2 \bmod{7}

    67\equiv 1 \bmod{2}
    67\equiv 1 \bmod{3}
    67\equiv 2 \bmod{5}
    67\equiv 4 \bmod{7}

    Now if you add ‘x’ to each of 23,29,31,37,67 to get a prime corresponding to each number, then

    x\equiv 0 \bmod{2} (otherwise, if you add ‘x’ such that x\equiv 1 \bmod{2}, then you will not get any prime)
    Similarly,
    x\equiv 0 \bmod{3}
    x\equiv 0 \bmod{5}
    and
    x\equiv 0 \bmod{7} or x\equiv 1 \bmod{7} or x\equiv 2 \bmod{7}

    Now consider the first case i.e. when
    x\equiv 0 \bmod{2}
    x\equiv 0 \bmod{3}
    x\equiv 0 \bmod{5}
    x\equiv 0 \bmod{7}, then the first solution to these four equations will be ‘x=210’.
    The general solutions to these four equations is given by ‘x=210n’ where n is a positive integer.
    Now, we can add ‘x=210n’ to each of 23,29,31,37,67,1949 & check whether we are getting all numbers to be prime or not. If we are getting all numbers to be prime then ‘1949+210n’ will be a golden year. The first three solution obtained in this way are 3209,4259,5939.

    Now consider the second case i.e. when
    x\equiv 0 \bmod{2}
    x\equiv 0 \bmod{3}
    x\equiv 0 \bmod{5}
    x\equiv 1 \bmod{7}, then the first solution to these four equations will be ‘x=120’.
    Now, using Chinese remainder theorem we get the general solutions to these four equations be ‘x=120+210n’ where n is a non-negative integer.
    Now, we can add ‘x=120+ 210n’ to each of 23,29,31,37,67,1949 & check whether we are getting all numbers to be prime or not. If we are getting all numbers to be prime then ‘1949+120+210n’ will be a golden year. One such golden year obtained is 3539.

    Now consider the second case i.e. when
    x\equiv 0 \bmod{2}
    x\equiv 0 \bmod{3}
    x\equiv 0 \bmod{5}
    x\equiv 2 \bmod{7}, then the first solution to these four equations will be ‘x=30’.
    Now, using Chinese remainder theorem we get the general solutions to these four equations be ‘x=30+210n’ where n is a non-negative integer.
    Now, we can add ‘x=30+ 210n’ to each of 23,29,31,37,67,1949 & check whether we are getting all numbers to be prime or not. If we are getting all numbers to be prime then ‘1949+30+210n’ will be a golden year. One such solution is 1979.

    Note that I have not used ‘modulo 11,13 or higher’ because they will provide a very little resolution as we are using only five numbers 23,29.31,37,67.

    This is exactly the way I got few solutions.

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  2. If you consider the ages of all persons to be positive only, then there was no golden year for family before 1949. The next golden year for the family will be 1979, & the ages of the persons will be 97, 67, 61, 59, 53 respectively. The numbers 1979, 97, 67, 61, 59, 53 are all primes.
    If you consider general solution to this problem, the other solutions, that is the other golden years for the family will be 3209, 3539, 4259, 5939. I think, this problem must have an infinite numbers of solutions, but I still have to prove it.
    One important point about the solutions to this problem is that whenever we get a golden year by adding a number ‘x’ to 1949, the number ‘x’ will must be of the form ‘210x’ (as for the years 3209, 4259, 5939) or ’30+210x’ (as for the year 1979) or ‘120+210x’ (as for the year 3539).

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    • Your answer and observation is correct. We can workout a proof for your claim if you please elaborate the method you followed to find the solutions.
      You can use \LaTeX in comments. [http://en.support.wordpress.com/latex/]

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