Posers and Puzzles

Posers and Puzzles

  1. Joined
    11 Jun '06
    Moves
    3516
    28 Jun '06 17:281 edit
    A large company has an annual new year's lottery. Every employee puts in a dollar bill then every emploee picks a number from 1 to 100. If your number matches the randomly generated number, you win.
    in the case of a tie the pot is split evenly and any remaining dollar bills go to charity.

    Year 1: 41 people split the pot charity gets $12
    Year 2: 19 people split the pot charity gets $7
    year 3: 37 people split the pot charity gets $31

    If the number of employees is constant what is the minimum number of employees at the company?

    what is the maximum number of employees if no employee has reached the maximum allowable winnings of $5000?
  2. Standard memberXanthosNZ
    Cancerous Bus Crash
    p^2.sin(phi)
    Joined
    06 Sep '04
    Moves
    25076
    28 Jun '06 17:442 edits
    Minimum number of employees is 5,137. Maximum number of employees with no prize exceeding $5000 is 91,606.

    I think.

    EDIT: For anyone who is interested I obtained my answers using a quickly written piece of code that searched for numbers that returned true to:

    if (mod(n,41) == 12) & (mod(n,19) == 7) & (mod(n,37) == 31)

    I'd be interested to see if a non-brute force solution exists given the one way nature of the modulus function.
  3. Joined
    11 Jun '06
    Moves
    3516
    28 Jun '06 19:04
    Originally posted by XanthosNZ
    Minimum number of employees is 5,137. Maximum number of employees with no prize exceeding $5000 is 91,606.

    I think.

    EDIT: For anyone who is interested I obtained my answers using a quickly written piece of code that searched for numbers that returned true to:

    if (mod(n,41) == 12) & (mod(n,19) == 7) & (mod(n,37) == 31)

    I'd be interested to see if a non-brute force solution exists given the one way nature of the modulus function.
    of course it does.
    i think posting an answer using a computer to get it isn't very fair to everyone else
  4. Joined
    11 Jun '06
    Moves
    3516
    28 Jun '06 19:13
    can you solve this one in your head?

    Year 1: 41 people split the pot charity gets $40
    Year 2: 19 people split the pot charity gets $18
    year 3: 37 people split the pot charity gets $36

    how about?

    Year 1: 41 people split the pot THE REMAINING $12 GET ADDED TO NEXT YEARS POT
    Year 2: 19 people split the pot ; $7 carries to year 3
    year 3: 37 people split the pot charity gets $31

    please no computer generated solutions.

    Describe all possible solutions to the original problem (computer can't help you here)
  5. Standard memberXanthosNZ
    Cancerous Bus Crash
    p^2.sin(phi)
    Joined
    06 Sep '04
    Moves
    25076
    28 Jun '06 19:32
    I'm an engineer, the easiest way of finding a solution is the best one.
  6. Joined
    11 Jun '06
    Moves
    3516
    28 Jun '06 19:44
    Originally posted by XanthosNZ
    I'm an engineer, the easiest way of finding a solution is the best one.
    find any solution to the simultaneous congruences

    x~387324mod 10^ 572
    x~24242357mod 3^428

    enjoy checking all the numbers until you get one that works
  7. Standard memberuzless
    The So Fist
    Voice of Reason
    Joined
    28 Mar '06
    Moves
    9908
    28 Jun '06 19:59
    Originally posted by aginis
    A large company has an annual new year's lottery. Every employee puts in a dollar bill then every emploee picks a number from 1 to 100. If your number matches the randomly generated number, you win.
    in the case of a tie the pot is split evenly and any remaining dollar bills go to charity.

    Year 1: 41 people split the pot charity gets $12
    Year 2: 19 peopl ...[text shortened]... maximum number of employees if no employee has reached the maximum allowable winnings of $5000?
    call me crazy but....

    The min number of employees is 26 if in year 2 19 people shared 1 dollar each and the remaining $7 went to charity.

    The max number would be 5000 if the max pot is $5000 since everyone can only put $1 dollar into the pot.
  8. Joined
    11 Jun '06
    Moves
    3516
    28 Jun '06 20:27
    Originally posted by uzless
    call me crazy but....

    The min number of employees is 26 if in year 2 19 people shared 1 dollar each and the remaining $7 went to charity.

    The max number would be 5000 if the max pot is $5000 since everyone can only put $1 dollar into the pot.
    if there are 26 employs how did 41 people split the pot in the first year? remember same number of employees means same pot size every year. (in the original version)

    the max prize is 5000 but if you split a $19,000 pot 19 ways each employee only recieves $1000. etc.
  9. B is for bye bye
    Joined
    09 Apr '06
    Moves
    27526
    28 Jun '06 21:53
    Originally posted by XanthosNZ
    I'm an engineer, the easiest way of finding a solution is the best one.
    I thought this site was supposed to be engineer free?
  10. Joined
    04 Jan '04
    Moves
    25350
    29 Jun '06 13:11
    Originally posted by aginis
    can you solve this one in your head?

    Year 1: 41 people split the pot charity gets $40
    Year 2: 19 people split the pot charity gets $18
    year 3: 37 people split the pot charity gets $36

    how about?

    Year 1: 41 people split the pot THE REMAINING $12 GET ADDED TO NEXT YEARS POT
    Year 2: 19 people split the pot ; $7 carries to year 3
    year 3: 37 people ...[text shortened]... ions.

    Describe all possible solutions to the original problem (computer can't help you here)
    Solutions to the original problem will be like this:

    n = 5137 + 28823x (where 28823 = 19 * 37 * 41)

    This gives us a clue to how e can solve the next problem "in our heads"

    General solutions will be of the form n = A + 28823x - where is any answer that works. Clearly -1 works (-1 mod 41 = 40, -1 mod 19 = 18 etc - which is why you chose those numbers), so the lowest, real world solution will be -1 + 28823 * 1 = 28822.

    I can't do the carries-over problem in my head though :-(
  11. Joined
    04 Jan '04
    Moves
    25350
    29 Jun '06 13:13
    Originally posted by XanthosNZ
    I'm an engineer, the easiest way of finding a solution is the best one.
    Have engineers stopped coding in FORTRAN then?
Back to Top