Can you guess the number of pirates and the share of each?
Source: Reader Sudipta Das.
The only proportions that satisfy the conditions bulleted in the problem statement are:
2 Pirates: 3:4
3 Pirates: 5:11:13
4 Pirates: 1:19:25:28
[Thus, any solution must be multiple of one of the above proportions.
The first statement "more than 50%..." rules out 2 pirates and restricts
the other proportions to odd multiples.
The second statement "The least ..." forces the multiple for the 4-pirate
proportion to be 5, 15, etc (so that the possible solutions both have the
same minimum value). If the multiple is greater than 13, then the
total number of coins is greater than 1000, so the multiple can only be 5.
The last statement "...more than 10 times..." forces the solution to be
that for 4-pirates.]
Given the rest of the conditions, the answer is:
4 pirates with shares 5, 95, 125, 140.
I have to say I found the end a bit confusing in terms of getting rid of the possible solutions. He's what I did, maybe I'll find my error on route :-) First I found out possible solutions with 3 pirates (getting a,b,c coins) since we know there must be at least three from the puzzle statement. The two pirate solution is clearly a=3, b-=4 but the puzzle seems has three equalities so I'm assuming 3 minimum. a+(b+c)/2 = b+ (a+c)/3 = c+(a+b)/4 as per statement. Resolving and re-arranging, I got 13a=5c. Therefore the smallest integer solution is a=5, c=13 (and so b=11) and all multiples thereof. Then I looked at four pirates : a+(b+c+d)/2 = b+(a+c+d)/3 = c+ (a+b+d)/4 = d+ (a+b+c)/5 Resolving and re-arranging I got 28c=25d. Therefore the smallest integar solution is c=25, d=28 giving a=1, b=19 and multiples thereof. Moving onto five pirates .. I have not been able to get an answer out. I have not tried higher pirates (this would be an algebraic nightmare!)
f we let X equal to the total amount of gold, and P1, P2, ... equal the shares of the pirates, then we can write the main equation as: 1/2 X + 1/2 P1 = 1/3 X + 2/3 P2 = ... 1/(N+1) X + N/(N+1) PN Taking consecutive pairs of terms from this, we get the following: 2/3 P2 - 1/2 P1 = 1/6 X or 4 P2 - 3 P1 = X 3/4 P3 - 2/3 P2 = 1/12 X or 9 P3 - 8 P2 = X ... Alternatively, we can write these as: P2 = (X + 3 P1)/4 P3 = (X + 8 P2)/9 ... If we substitute the expression for P2 into the expression for P3, we get: P3 = 1/9 X + 8/9 P2 = 1/9 X + 2/9 X + 2/3 P1 = 1/3 X + 2/3 P1 Similarly, for other PN: P4 = 3/8 X + 5/8 P1 P5 = 2/5 X + 3/5 P1 P6 = 5/12 X + 7/12 P1 ... PN = (N-1)/(2N) X + (N+1)/(2N) P1 Now, if we assume there are two pirates: X = P1 + P2 = P1 + 1/4 X + 3/4 P1 = 1/4 X + 7/4 P1 3/4 X = 7/4 P1 P1 = 3/7 X P2 = 4/7 X If there are three pirates: X = P1 + P2 + P3 = P1 + 1/4 X + 3/4 P1 + 1/3 X + 2/3 P1 = 7/12 X + 29/12 P1 5/12 X = 29/12 P1 P1 = 5/29 X P2 = 11/29 X P3 = 13/29 X If there are four pirates: X = P1 + P2 + P3 + P4 = P1 + 1/4 X + 3/4 P1 + 1/3 X + 2/3 P1 + 3/8 X + 5/8 P1 = 23/24 X + 73/24 P1 1/24 X = 73/24 P1 P1 = 1/73 X P2 = 19/73 X P3 = 25/73 X P4 = 28/73 X If there are more than four pirates, then the similar algebra to the above gives X = (greater than 1) X + (positive) P1 which gives P1 is negative. Thus there are no more than 4 pirates. The clue "more than 50% of us received an odd number of coins" eliminates the possibility of there being only 2 pirates, because the one who received 4/7 of the coins would receive an even number of coins. At this point, Jolly Rogers can deduce that there are either 3 pirates, who received numbers of coins 5, 11, and 13 times an odd number from 1 to 33, or there are 4 pirates, who received numbers of coins 1, 19, 25, and 28 times an odd number of coins from 1 to 13. When Jolly Roger is told the least number of coins a pirate got, if that number was 15, 25, 35, ... then he'd know there were three pirates, and the numbers of coins each pirate got. If it was 1, 3, 7, 9, 11, or 13, then he'd know there were four pirates. Only if the fewest coins was 5 would he still not know. In order for one pirate to get 10 times what another one got, there must be 4 pirates, and the multiplier is 5. The pirates therefore got 5, 95, 125, and 140 coins.