Ron Barnette's Zeno's Coffeehouse Challenge #51 Result
Dear Zeno's patrons: Here are the results of our previous challenge. I have listed the original problem, with the winning results listed below. Thanks to all Zeno's patrons who took up the challenge! Over 200 entries were received, and I have included results and analyses from six entries. Thanks for your critical thinking, and look to Zeno's next Challenge:)
As Maggie and Charles shared some time with long-standing Zeno's patrons a while
back, they reflected together on the wonderful ten-plus years that the Coffeehouse has
been operating for world travelers who have come to enjoy the mental gymnastics. To the
amazement of all, Maggie recounted a problem suggested by a patron who submitted a reply
to the fourth Zeno's Challenge, 'Maggie's Bet,' back in 1995! [Available on the website,
btw..RB]. Here is the problem that Maggie mentioned, from a patron's comments back way
back then. But is it solvable, as claimed?
Here's a problem for you: 3 thieves, A, B and C steal some money. They each take a handful of cash. When they count what they have taken, B and C complain to A that he has taken most. A replies that she will give B the same amount that B already has, and the same for C. Now they count the money again and now B has most so he does the same for A and C. But now C has most so C does the same for A and B. Finally they all end up with the same amounts. How much did they all start with? Give your answer in the simplest possible ratio.... Thanks for a very interesting site.
Emails to this Zeno's Coffeehouse patron proved unsuccessful. So Maggie shrugged and smiled, while, this time, Charles began calculating....Is there a solution?
Many Zeno's patrons thought that there was a solution, and several didn't. Here are some positive responses from six international patrons, whose detailed comments I include. Thanks goes to Jeffrey Boone, Andrew, Troy Williamson, Melanie Cannon, Topi Linkala, and Mark Young. Nice work! ...Ron Barnette
1. Jeffrey Boone, Loxley, AL, USA
comments: Ok, I enjoyed this one too.
A stole $13, B stole $7, and C stole $4 for a total robbage of $24, and all have $8 in the end (with keeping the same ratio, any multiples of this will work).
Here's how I got it:
A stole a number a.
B stole a number b.
C stole a number c.
When A gives to B and C,
A has a-b-c,
B has 2b, and
C has 2c.
When B gives to A and C,
A has 2a-2b-2c,
B has 2b-(a-b-c)-2c, which simplifies to -a+3b-c, and
C has 4c.
When C gives to A and B,
A has 4a-4b-4c,
B has -2a+6b-2c, and
C has 4c-(2a-2b-2c)-(3b-a-c), which simplifies to -a-b+7c.
Since at this point all three numbers are equal,
Using matrices to solve this system of equations,
Simplifying to integers, the solution set is:
creating the ratio from a to c being:
showing that A stole 13, B stole 7, and C stole 4.
Thanks for another fun one!
2.Andrew,Worcester, MA, USA
comments: The answer is 13:7:4.
How did I get this? Well, lets assume that A starts with amount of money z, B with y, and C with x.
Then, if A gives money to B and C to double their amounts then A with have z-y-x, B will have 2y and C will have 2x.
Now, when B gives A and C money to double their amounts then A will have 2z-2y-2x, B will have 3y-x-z, and C will have 4x.
Now, C gives A and B money to double their amounts, and thus A has 4z-4y-4x, B has 6y-2x-2z, and C has 7x-y-z.
Now, we know these 3 amounts are equal. So setting A and C's amounts equal to each other (4z-4y-4x = 7x-y-z) and re-arranging yields 5z-3y-11x=0 (*).
Similarly, setting B and C's amounts equal to each other yields 7y-9x-z = 0 (**), or z = 7y-9x
substituting this into equation (*) gives us 32y-56x=0 which simplifies to y=(7/4)x
Now, multiplying equation (*) by 7 and equation (**) by 3 and adding the two together yields 32z-104x=0 or in other words z=(13/4)x
So we have z=(13/4)x and y=(7/4)x, so z,y and x have a ratio of (13/4):(7/4):1, or to simplify, a ratio of 13:7:4.
3. Troy Williamson, Abilene, TX, USA
comments: "A" began with 13/24 of the loot, "B" started with 7/24 of the loot, and "C" began with 1/6 of the loot.
I let "a" represent the portion of the loot with which "A" began, "b" the portion with which "B" began, and "c" the portion that "C" initially held. If "x" is the total amount stolen, then
ax + bx + cx = x
Dividing out the "x", I know that (using this setup)
a + b + c = 1
Call this Eq. 1.
At the beginning, they have
Now "A" is going to match the money held by "B" and "C".
A: a - b - c
B: b + b = 2b
C: c + c = 2c
Next, "B" matches the money held by "A" and "C".
A: (a - b - c) + (a - b - c) = 2a - 2b - 2c
B: 2b - (a - b - c)- 2c = -a + 3b - c
C: 2c + 2c = 4c
Finally, "C" matches the funds held by "A" and "B".
A: (2a - 2b - 2c) + (2a - 2b - 2c) = 4a - 4b - 4c
B: (-a + 3b - c) + (-a + 3b - c) = -2a + 6b - 2c
C: 4c - (2a - 2b - 2c) - (-a + 3b - c) = -a - b + 7c
I know that these final amounts held must equal each other.
Taking the final holdings of "A" and "B":
4a - 4b - 4c = -2a + 6b - 2a
6a - 10b - 2c = 0
3a - 5b - c = 0
Call this Eq. 2.
Next, take the final amounts held by "B" and "C":
-2a + 6b - 2c = -a - b + 7c
-a + 7b - 9c = 0
Call this Eq. 3.
I now have three equations with three variables.
Eq. 1: a + b + c = 1
Eq. 2: 3a - 5b - c = 0
Eq. 3: -a + 7b - 9c = 0
When these are solved (I used an augmented matrix), you get the result: a = 13/24, b = 7/24, and c = 1/6.
Check the result. At the outset,
After "A" distributes funds to "B" and "C":
A: 13/24 - 7/24 - 1/6 = 1/12
B: 7/24 + 7/24 = 7/12
C: 1/6 + 1/6 = 1/3
After "B" distributes funds to "A" and "C":
A: 1/12 + 1/12 = 1/6
B: 7/12 - 1/12 - 1/3 = 1/6
C: 1/3 + 1/3 = 2/3
After "C" distributes funds to "A" and "B":
A: 1/6 + 1/6 = 1/3
B: 1/6 + 1/6 = 1/3
C: 2/3 - 1/6 - 1/6 = 1/3
Hence, whatever the amount of loot stolen, "A" began with 13/24 of it, "B" began with 7/24 of it, and "C" began with 1/6 of it (implying that the amount stolen can be evenly divided into 24 equal parts!); in the end, all three thieves held 1/3 of the stolen loot.
4. Melanie Cannon, New Westminster, BC, Canada
comments: Hi Ron. I've been playing around with this problem for awhile and this is the best answer I've been able to come up with:
A starts off with $12448.34, B starts off with 6702.95, and C starts off with 3830.26
A: $12448.34 - $6702.95 - $3830.26 = $1915.13
B: $6702.95 * 2 = $13405.90
C: $3830.26 * 2 = $7660.52
A: 1915.15 * 2 = 3830.26
B: 13405.90 - 1915.13 - 7660.52 = 3830.25
C: 7660.52 * 2 = 15321.04
Round 3 (END RESULT)
A: 3830.26 * 2 = $7660.52
B: 3830.25 * 2 = $7660.50
C: 15321.04 - 3830.25 - 3830.26 = $7660.53
I suppose you could get the amounts a bit closer by moving the decimal point around, but at the moment I can't find a way for them to be exact. Perhaps A and C dropped a couple pennies on the way to the get-away car.
Thanks again for a great site!
further comments: Okay, I think I've finally got it: the smallest whole number ratio is
Beginning: A has 13, B has 7, C has 4
Round 1: A has the most money and gives 7 to B and 4 to C. Now A has 2, B has 14, and C has 8
Round 2: Now B has the most and gives 2 to A and 8 to C. So, A has 4, B has 4, and C has 16
Round 3: C has the most, so he gives 4 to A and 4 to B. Now A has 8, B has 8, and C has 8.
5. Topi Linkala, Helsinki, Finland
comments: Sorry that I left the calculations from my previous answer. Here they are.
A B C
Start a b c
1st exchange a-b-c 2b 2c
2nd exchange 2a-2b-2c 2b-(a-b-c)-2c 4c
3rd exchange 4a-4b-4c -2a+6b-2c 4c-(2a-2b-2c)
So we need to solve the two equations:
1) 4a-4b-4c = -2a+6b-2c and
2) 4a-4b-4c = -a-b+7c
4a-4b-4c = -2a+6b-2c <=>
4a+2a = 6b+4b-2c+4c <=>
6a = 10b+2c <=>
3a = 5b+c (1)
4a-4b-4c = -a-b+7c <=>
4a+a = -b+4b+7c+4c <=>
5a =3b+11c (2)
From (1) and (2) we get the following:
(1') 15a = 25b+5c and
(2') 15a = 9b+33c
And from these:
25b+5c = 9b+33c <=>
25b-9b = 33c-5c <=>
16b = 28c <=>
4b = 7c (3)
So by trying c=4 we get:
c = 4
4b = 7*4 (3)
b = 7
3a = 5*7 + 4 (1)
3a = 39
a = 13
6. Mark Young, Wolfville, NS, Canada
comments: Let a be the amount A got, b the amount B got, and c the amount C got.
Each exchange doubles the money of the two poorer criminals. A is the richest at first, so after the first exchange of money:
A has a - b - c
B has 2b
C has 2c
Then B is the richest, so after the second exchange
A has 2(a - b - c) = 2a - 2b - 2c
B has 2b - (a-b-c) - 2c = -a + 3b - c
C has 2(2c) = 4c
Then C is the richest, so after the third exchange
A has 2(2a - 2b - 2c) = 4a - 4b - 4c
B has 2(-a + 3b - c) = -2a + 6b - 2c
C has 4c - (2a-2b-2c) - (-a+3b-c) = -a - b + 7c
We know that they eventually ended up with equal amounts, so we know that a+b+c is divisible by three. Let's guess that the amount is also divisible by 4 and 5 -- hoping to keep the math easy! (We could use another variable, but it's easier to see mistakes if we use an actual number.)
So, the total amount is (we guess) $60.
(1) 1a + 1b + 1c = 60
Since each thief ended up with 20 (we guess), we also have
(2) 4a - 4b - 4c = 20
(3) -2a + 6b - 2c = 20
(4) -1a - 1b + 7c = 20
We can add (1) and (4) together:
(5) 8c = 80
So C started with $10. Substitute that into (1) and (2)
(6) 1a + 1b = 60 - 10 = 50
(7) 4a - 4b = 20 + 40 = 60
Multiply (6) by 4
(8) 4a + 4b = 200
Add that to (7)
(9) 8a = 260
So A started with 260/8 = $32.50, meaning that B started with $50 - $32.50 = $17.50.
We double those amounts to get the smallest /integer/ ratio (dang, we guessed too small), and we find that
A started with $65,
B started with $35, and
C started with $20.
We check our answer by showing how the amounts changed:
A B C
-- -- --
65 35 20
10 70 40
20 20 80
40 40 40
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