Puzzle #154: Can You Divvy Up the Gold?

I continue my love affair with the NSA puzzles, and here is one more, third week in a row – thanks again Suman!

We have done a few pirates and gold problems in the past. This one is slightly easier. It is a good puzzle to try with middle graders For adults/ avid puzzlers, this will be more of a speed test!

Following their latest trip, the 13 pirates of the ship, SIGINTIA, gather at their favorite tavern to discuss how to divvy up their plunder of gold coins. After much debate, Captain Code Breaker says, “Argggg, it must be evenly distributed amongst all of us. Argggg.” Hence, the captain begins to pass out the coins one by one as each pirate anxiously awaits her reward. However, when the captain gets close to the end of the pile, she realizes there are three extra coins.

After a brief silence, one of the pirates says, “I deserve an extra coin because I loaded the ship while the rest of you slept.” Another pirate states, “Well, I should have an extra coin because I did all the cooking.” Eventually, a brawl ensues over who should get the remaining three coins. The tavern keeper, annoyed by the chaos, kicks out a pirate who has broken a table and who is forced to return her coins. Then the tavern owner yells, “Keep the peace or all of you must go!”

The pirates return to their seats and the captain, left with only 12 total pirates, continues to distribute the coins – “one for you,” “one for you.” Now, as the pile is almost depleted, she realizes that there are five extra coins. Immediately, the pirates again argue over the five extra coins. The captain, fearing that they will be kicked out, grabs the angriest pirate and ushers her out of the tavern with no compensation. With only 11 pirates left, she resumes distribution. As the pile nears depletion, she sees that there won’t be any extra coins. The captain breathes a sigh of relief. No arguments occur and everyone goes to bed in peace.

If there were less than 1,000 coins, how many did the pirates have to divvy up?

As always, please send your answers as comments within the blog (preferred), or send an e-mail to alokgoyal_2001@yahoo.com. Please do share the puzzle with others if you like, and please also send puzzles that you have come across that you think I can share in this blog.

Happy pirating!

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8 Responses to Puzzle #154: Can You Divvy Up the Gold?

  1. mashex says:

    The Answer is 341 .With the help of some simple login and python (programming language )it was very easy to figure it out.
    The logic is that when the number is divided by 11 ,it should have no remainder as the gold is evenly distributed .Similarly when divided by 12 and 13 ,a remainder of 5 and 3 should come .
    So the only number between 1 and 1000 ,which met this condition was the number 341.

    The Code is as follows:
    for num in range(1,1000):
    if num%11==0 and num%12==5 and num%13==3:
    print num
    %-For getting remainder

  2. Amit Mittal says:

    Agree, kids should have fun with that. Still, having spent some time on its advanced forms, here is my solution:

    I struck speed after attempting basic congruent math ( ax = b*mod[c] => gcd [a,b] exists, but soon found a root 653 which satisfies 3 mod 13 and 5 mod 12, also 13 being prime other roots that satisfy the first two conditions work to a difference of 156 apart (12*13) . that leaves me testing 653,809,965 and on the lower side 497,341 and 185 . Elegantly we look at all multiples of 156 and get where the numbers are 3+13x and 5+12 y, to the series:

    185,341,497,653,809 and 965 and testshows only 341 is divisible in 11 pirates. Enjoy the reunion at IIT Delhi!

  3. Abhinav Jain says:

    341. Clearly one has to figure out the smallest no. Which gives remainder 3 with 13 , 5 with 12 and 0 with 11.with simple hit and trial it can be figured out

  4. Praneeth says:

    11z = 12y+5 = 13x+3 , so y+5 and 2x+3 should be divisible by 11 but we knew that the max value of y is 72 because 12y+5 < 1000 and also y can take values 6, 17, 28, 39, 50, 61 and 72. Finally checking these numbers gives y = 28 that satisfies all the conditions and the number of coins are 341 = 12*28 + 5.

  5. Let the number of coins be n.

    n mod 13 = 3, n mod 12 = 5, n mod 11 = 0

    Say n = 12x + 5, n x y y = 2
    => x = 28
    n = 341

  6. Some problem with the last comment I put.

    Let the number of coins be n.

    n mod 13 = 3, n mod 12 = 5, n mod 11 = 0

    Say n = 12x + 5, n is less than 1000
    => x is less than 85

    n mod 13 = 5-x
    x mod 13 = 2

    n mod 11 = x+5
    x mod 11 = 6

    Say x = 13y + 2
    => y is less than 8

    2y+2 mod 11 = 6
    2y mod 11 = 4

    => y = 2
    => x = 28
    n = 341

  7. Batman says:

    We have

    11a=12b+5=13c+3. Where a,b,c are natural numbers.

    Let us first look at at the equation 12b+5=13c+3 => 12b-13c+2=0.It is not very difficult to see that b=c=2 is a solution of the equation(although one can use the euclidean algorithm if required). Hence we can write the general solution of this equation as

    12(2+13t) – 13(2+12t) +2=0. Where t is a natural number.

    Hitting trial and error by putting different values of t we need to find a solution for which 12b+5=12(2+13t)+5 is a multiple of 11. Seeing that t<7(since 12(2+13t)+5<1000), it doesn't seem like an reasonable task of doing trial and error.

    Putting t=2 we get that 12*28+5=341 divides 11. Hence we get 341 as our answer.

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