Gold Bars in a Pile
Three robbers A, B, and C each place stolen gold bars in a common pile, their shares being 1/2, 1/3, and 1/6 of the total amount, respectively. Next, each man takes the bars from the pile until none remain. Now A returns 1/2 of what he took, B 1/3, and C 1/6. When the last pile of gold is divided equally by the three men, it turns out that each has the amount he had at the beginning.
What was the total number of gold bars? (look for the minimum number of gold bars to satisfy the conditions)
What was the total number of gold bars? (look for the minimum number of gold bars to satisfy the conditions)
Labels: mathschallenge
posted by Zaux at
10:57 PM
24 Comments:
$5922 or any multiple thereof.
I'm probably not reading the question right, but the last sentence of the first para. "it turns out that each has the amount he had at the beginning, before removing money from the pile" confuses me. They have just put all their money into the pile? They would have no money left....? Leading me to the answer 0, and 3 guys practicing an exercise in futility?
Is the answer in £, $, yen or Euros? (if Euros, I will refuse to try to answer the question...)
Hi Karl...
(last sentence, first par.):
When the men equally divided the money in the second pile, they then each had the amount of money they had in the beginning ...after reading the problem a dozen times, I agree with you ... I will alter the statement
Thanks Karl ... wording of the problem statement was not mine ... but I changed it for clarity.
Money in a pile
The wording is killing me on this... I must be interpreting it wrong
Interpretation A)
1. All money goes in pile
2. All money then gets split equally i.e. 1/3
3. A,B,C returns 1/2,1/3,1/6 of their equal share into a second pile
4. The second pile is divided up equally i.e. 1/3
5. A,B,C are left with 1/2 of the original pile, 1/3 of the original pile, 1/6 of the original pile
Interpretation A is impossible for all but 0 as:
Call money in original pile S
After Step 2
A = 1/3* S, B=1/3*S, C= 1/3*S
After Step 3
A=1/3*S-1/2*1/3*S=(1/3-1/6)*S=1/6*S
B=1/3*S-1/3*1/3*S=(1/3-1/9)*S=2/9*S
C=1/3*S-1/3*1/6*S=(1/3-1/18)*S=5/18*S
2nd pile=(1/6+1/9+1/18)*S=1/3*S
After Step 4
1/3 of 2nd pile=1/9*S
A=(1/6+1/9)*S=5/18*S
B=(2/9+1/9)*S=1/3*S
C=(5/18+1/9)*S=7/18*S
Statement 5
A=5/18*S=1/2*S
B=1/3*S=1/3*S
C=7/18*S=1/6*S
from C
7/18*S=1/6*S
4/18*S=0
S=0
Answer for this interpretation:
S can only be 0
Please identify which part I am not interpreting as intended
Cam
Hopefully clarification:
* 3 men place their money in a pile.
* of the pile money,1/2 came from A, 1/3 came from B, 1/6 came from C
* Now the men take all the money, each taking a different amount
* Then A returns to the pile 1/2 of what he took
* B returns 1/3 of what he took.
* C returns 1/3 of what he took.
* Now the new pile of money is equally divided by the 3 men.
* by totaling the money each had, the money each received when the last pile was divided ... and then adding each man's money, you get the same amount as in the original pile.
Money in a pile
Let's see if this interpretation works....
Interpretation B)
1. All money goes in pile
2. All take differnt amounts A takes X, B takes Y, C takes Z
3. A,B,C returns 1/2,1/3,1/6 of the amount they took (X,Y,Z) into a second pile
4. The second pile is divided up equally i.e. 1/3
5. A,B,C are left with 1/2 of the original pile, 1/3 of the original pile, 1/6 of the original pile
Call money in original pile S
After Step 2
A = X* S, B=Y*S, C= Z*S
After Step 3
A=X*S-X*1/2*S=1/2*X*S
B=Y*S-Y/3*S=2/3*Y*S
C=Z*S-1/6*Z*S=5/6*Z*S
2nd pile=(1/2*X+1/3*Y+1/6*Z)*S
After Step 4
1/3 of 2nd pile=
=1/3*(1/2*X+1/3*Y+1/6*Z)*S
=(1/6*X+1/9*Y+1/18*Z)*S
A= 1/2*X*S+(1/6*X+1/9*Y+1/18*Z)*S
B= 2/3*Y*S+(1/6*X+1/9*Y+1/18*Z)*S
C= 5/6*Z*S+(1/6*X+1/9*Y+1/18*Z)*S
Statement 5
A= 1/2*X*S+(1/6*X+1/9*Y+1/18*Z)*S=1/2*S
B= 2/3*Y*S+(1/6*X+1/9*Y+1/18*Z)*S=1/3*S
C= 5/6*Z*S+(1/6*X+1/9*Y+1/18*Z)*S=1/6*S
Yields 3 equations
1/18*(12X+2Y+1Z)=1/2
1/18*(3X+14Y+1Z)=1/3
1/18*(3X+2Y+16Z)=1/6
OR
*1) 12X+2Y+1Z=9
*2) 3X+14Y+1Z=6
*3) 3X+2Y+16Z=3
From *1-*3
9X-15Z=6
3X=2+5Z
*4) X=2/3+5/3*Z
Sub *4 into *1
8+20Z+2Y+1Z=9
2Y+21Z=1
*5) Y=1/2-21/2*Z
Sub *4 and *5 into *2
2+5Z+7-147Z+1Z=6
9-141Z=6
141Z=3
*5) Z=3/141=9/423=18/846
X=2/3+5/141=297/423=594/846
Y=1/2-21/2*3/141=1/2-63/282=78/282=234/846
X=18/846
Y=594/846
Z=234/846
Reduce to
X=9/423
Y=297/423
Z=117/423
If A,B,C take the fractions X,Y,Z of S then it can always be done i.e. it is not dependent on S (the size of the original pile of money)
If, however, one puts a restraint saying that they must take whole numbers then, any multiple of 423 should work.
Cam
This latest stab seems to line up with Wiz, in that 5922 is a multiple of 423.(5922/423=14)
Cam
X=9/423 reduces to 1/47
Y=297/423 reduces to 33/47
Z= 117/423 reduces to 13/47
So multiples of 47 should work.
Cam
Hi Cam,
Your answer of 423 means that A gets a fractional amount in his share, i.e. 211.5.
How about we agree on 846 as the answer!
No one has posted the correct answer yet ... does everyone want to continue the pursuit ... or would you like the answer (and attempt to duplicate it), or would you like a complete solution?
Wiz,
You are, of course, right. Multiples of 846 it is.
Cam
Cam ... according to the source of the problem, the correct answer is a multiple of 47.
Zaux,
If that's not the right answer, I have to chalk it up to interpretation. Please submit the answer or identify where we have misinterpreted the problem. Otherwise I don't see how we can explore the problem further.
Cam
Zaux,
Lol, lot's of cross posting here.
So is the answer any multiple of 47 or a specific multiple of 47 ?
Make it easy on us, and just submit the answer.....
Cam
The problem source states the answer as 282 ... do you guys want to work toward that ... or simply see the solution?
Meh...
multiples of 282
Cam
Zaux,
While you posted, I cross posted it must be multiples of 282 (don't believe it is only 282), unless the problem specified what is the smallest amount of money possible for them to accomplish what was described.
X=9/423 reduces to 1/47
Y=297/423 reduces to 33/47
Z= 117/423 reduces to 13/47
but we must divide X by 2,Y by 3, and Z by 6 and get whole numbers for all
so 47*6=282
Go ahead and post solution
Cam
Cam ... you are correct ... at the end of the lengthy solution to the problem, it is stated that 282 is the minimum sum which satisfies the conditions. Sorry 'bout that... the problem statement should have asked for the minimum amount to satisfy the problem. I should have read to the end of the solution and corrected the problem statement.
I just checked my figures. 282 is correct as the minimum solution.
All multiples, including my earlier 5922, are, of course, also valid.
Zaux,
Thanks for the problem. We definitely appreciate the effort you go through in posting them. The wording on this problem,however, gave us a lot of grief. May I suggest that you revise the wording on the problem to what I have in quotes below.
It would allow others who want to tackle the problem to avoid the issues that we came across.
Cam
"After a robbery, three men A, B, and C each place gold bars in a common pile. A brought in 1/2 of the gold bars, B brought in 1/3 of the gold bars, and C brought in 1/6 of the gold bars. The robbers then divided up the common pile in a manner that gave each of A,B, and C different amounts. Following this, A returned 1/2 of what he took, B 1/3, and C 1/6 into a second pile. The second pile was then divided equally by the three robbers. It turns out that, each robber now had the number of gold bars that they had each brought in at the beginning.
What is the smallest possible amount of gold bars in the original pile?
(Assume all tranactions must take place with a whole number of gold bars, as they can not be cut up)"
Let me add my voice to the chorus of appreciation for you efforts, Zaux, hope you can keep it up!
I personally had no difficulty with the wording of this problem, and I always assumed that you were looking for the minimum solution.
There have been other puzzles in the past, from all contributors, which have been rather ambiguous. However, I do appreciate that it must be dificult to anticipate in advance what interpretations others may place on your words, so that clarifications may often be needed.
Happy New Year to all.
Again from me, I appreciate the postings - and I think the different interpretations of the questions, showing different approaches are what make this site so much more facinating that a simple Q&A site...
Thanks for the good words guys ... I love puzzles, games, and competition of any nature. My math skills are reasonalble, but certainly not at the the level of some of you guys.
I enjoy trying to find problems which will be fun and still very challenging. Some of the problem wordings are the not the best ... I often change them for clarity... but as one of you guys stated, the different interpretations also add additional interest.
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