Rich man’s and gold Coin puzzle

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A rich man died. In his will, he has divided his gold coins among his 5 sons, 5 daughters and a manager.

According to his will: First give one coin to manager. 1/5th of the remaining to the elder son. Now give one coin to the manager and 1/5th of the remaining to second son and so on….. After giving coins to 5th son, divided the remaining coins among five daughters equally.

All should get full coins. Find the minimum number of coins he has?

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  • 1 Answer(s)

    This is an interesting problem.

    We know that if the fifth son gets x5 then the fourth son gets x4 = 5*(x5 + 1)/4 – 1 = x5 + (x5+1)/4 and third son gets x3 = 5*(x4 + 1)/4 – 1 = x4 + (x4+1)/4 and so on.  Representing what each son gets in terms of what the fifth son gets gives:

    x3 = x5 + (x5+1)/4 + ((x5+(x5+1)/4))/4
    = x5 + (x5+1)/4 + 5(x5+1)/16
    x2 = x5 + (x5+1)/4 + 5(x5+1)/16 + 25(x5+1)/64
    x1 = x5 + (x5+1)/4 + 5(x5+1)/16 + 25(x5+1)/64 + 125(x5+1)/256

    From this we know that the fifth son (x5) must get a number of coins that can be expressed in the form x5 = 256y – 1.

    The other requirement is that x5 must be a multiple of 5, because each daughter gets 4/5 of what the 5th son gets.  That means that the final digit of y must be either 1 or 6 so that the final digit of 256y is 6.  The smallest such value for y is 1, making x5 = 255.

    Plugging in 255 for x5 gives

    x1 = 624
    x2 = 499
    x3 = 399
    x4 = 319
    x5 = 255
    with each daughter getting 4/5 * 255 = 204 and the manager getting 5 gold coins

    204 * 5 + 255 + 319 + 399 + 499 + 624 + 5 = 3121 = 624 * 5 + 1.  The minimum number of coins is 3121.

    The second smallest number of coins possible is 18,746, which results from setting y = 6.

    dougbell Genius Answered on 20th October 2015.
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