You may also like

problem icon

Double Time

Crack this code which depends on taking pairs of letters and using two simultaneous relations and modulus arithmetic to encode the message.

problem icon

Real(ly) Numbers

If x, y and z are real numbers such that: x + y + z = 5 and xy + yz + zx = 3. What is the largest value that any of the numbers can have?

problem icon

Overturning Fracsum

Solve the system of equations to find the values of x, y and z: xy/(x+y)=1/2, yz/(y+z)=1/3, zx/(z+x)=1/7

Leonardo's Problem

Age 14 to 18 Challenge Level:

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 $6$.