Leonardo's Problem
Problem
Three people (Alan, Ben and Chris), collectively own a certain number of gold coins. Respectively they own a half, one third and one sixth of the total number.
All the coins were piled on a table and each of them grabbed a part of the pile so that none were left.
After a short time:
Alan returned half of the coins that he had taken.
Ben returned a third of the coins that he had taken.
Chris returned one sixth of the coins that she had taken.
Finally each of the three got an equal share of the amount that had been returned to the table.
Surprisingly, each person had exactly the number of coins that really belonged to them.
What is the smallest number of coins that this strange transaction will work for?
How much did each person grab from the pile?
Getting Started
Student Solutions
Strangely, a whole 3 months after this problem first appeared, and all within a couple of days, four solutions came in from four different parts of the world. They were all excellent solutions, so well done Mehmet of Robert College, Turkey; Bradley of Avery Coonley School, Downers Grove, USA; Ling of Tao Nan School, Singapore and Edwinof The Leventhorpe School, Sawbridgeworth, England. We have re-produced Edwin's solution in full below.
To find the solution to this problem I started by finding expressions for the amount owned by Alan, Ben and Chris ($A$, $B$ and $C$ respectively) in terms of the total number ($T$). This gave:
$A = T / 2$
$B = T / 3$
$C = T / 6$
So $A = 3B / 2$
$A = 3C$
$B = 2C$
I then found how many were returned to the table, in terms of the number they each grabbed ($a$ , $b$ and $g$ respectively):
$(a /2 + b /3 + g /6) = (3a +2b + g ) / 6$
Then $A$, $B$ and $C$ in terms of $a$ , $b$ and $g$ :
$A = a / 2 + (3a +2b + g )/ 18$
$B = 2b / 3 + (3a +2b + g )/ 18$
$C = 5g / 6 + (3a +2b + g )/ 18$
( where $(3a +2b + g ) /18$ is an equal share of the amount returned to the table )
$A = (12a + 2b + g ) /18$
$B = (3a +14b + g ) / 18$
$C = (3a +2b + 16g ) /18$
Knowing the relationships between $A$ and $B$; $A$ and $C$; and $B$ and $C$, I found the simultaneous equations:
$15a - 38b - g = 0$
$3a - 4b - 47g = 0$
$3a - 10b + 31g = 0$
These did not have a unique solution but gave:
$b = 13g$
$a = 33g$
Putting these into the expressions for $A$, $B$, $C$ and $T$ gave:
$A = 47g / 2$
$B = 47g / 3$
$C = 47g / 6$
$T = 47g$
The lowest value of $g$ that these will give a whole number of sovereigns for is $6$, in which case:
$T = 282$
$A = 141$
$B = 94$
$C = 47$
$a = 198$
$b = 78$
$g = 6$
So Alan grabbed $198$, Ben grabbed $78$ and Chris grabbed $6$.
Teachers' Resources
Why do this problem?
The problem requires interpretiation of the information given and the creation of equations from that information. Discussion of the ways of interpreting the information could help to develop team working and communication skills.Possible approach
You might give learners some time to work independently on this problem then get them to share their ideas in small groups or pairs.Key questions
What are the unknowns?What equations an we write down based on the information given?