Tweedle dum and Tweedle Dee
Problem
They were standing under a tree, each with an arm round the other's neck, and Alice knew which was which in a moment, because one of them had "DUM" embroidered on his collar, and the other "DEE."
"I suppose they've each got 'TWEEDLE' round at the back of the collar," she said to herself.
Two brothers were left some money, amounting to an exact number of pounds all in coins, to divide between them. DEE undertook the division.
"But your heap is larger than mine!" cried DUM.
"True," said DEE. "Allow me to present you with one-third of my heap."
DUM added it to his heap, and after looking thoughtfully at the now gigantic pyramid, he suddenly exclaimed "I am well off now! Here is half of the heap for you."
"You are generous," said DEE, as he swept up the money. "Two-thirds of this heap is the least I can offer you."
"I will not be outdone in generosity!" cried DUM, hastily handing over three-quarters of his property.
"Prudence is a virtue," remarked DEE. "Content yourself with two-thirds of my present wealth."
"One-third of mine is all I can now afford!" retorted DUM.
"And now, if I give you one pound," remarked DEE, "we shall, I think, be square?"
He was right. How much money was shared between them?
Student Solutions
There are many solutions to this problem. Here is one correct solution received from a student at West Flegg Middle School in Norfolk.
| Dee | Dum |
| 48 | 28 |
| 32 | 44 |
| 54 | 22 |
| 18 | 58 |
| 61.50 | 14.50 |
| 20.50 | 55.50 |
| 39 | 37 |
| 38 | 38 |
Hence, they share 76 pounds between them.
Year 5 pupils at Sutton High School also found a couple of solutions.
Ewan, from King Edward VII School in Sheffield, used algebra to make sure he found all the possible solutions.
Let DUM begin with $x$ and DEE with $y$ coins. Note that $x < y$.
Following the giving of money algebraically the table shows DUM and DEE's fortunes after each gift:
| DUM | DEE |
| $x$ | $y$ |
| $x+\frac{1}{3}y$ | $\frac{2}{3}y$ |
| $\frac{1}{2}x+\frac{1}{6}y$ | $\frac{1}{2}x+\frac{5}{6}y$ |
| $\frac{5}{6}x+\frac{13}{8}y$ | $\frac{1}{6}x+\frac{5}{8}y$ |
| $\frac{5}{24}x+\frac{13}{72}y$ | $\frac{19}{24}x+\frac{59}{72}y$ |
| $\frac{53}{72}x+\frac{157}{216}y$ | $\frac{19}{72}x+\frac{59}{216}y$ |
| $\frac{53}{108}x+\frac{157}{324}y$ | $\frac{55}{108}x+\frac{167}{324}y$ |
After the last 'gift' of $1$ coin, we know that the totals are equal, so
\begin{align}
\frac{53}{108}x+\frac{157}{324}y+1 &= \frac{55}{108}x+\frac{167}{324}y -1 \\
2 &= \frac{2}{108}x+\frac{5}{162}y \\
324 &= 3x + 5y
\end{align}
Whole-number solutions of this equation are found by noticing that since $3x$ and $5y$ appear, if we introduce $n$ such that $x = 5n+a$ and $y=-3n+b$ then $n$ will cancel from the equation so we may find values for $a$ and $b$. One such solution is
\begin{align}
x &= 5n+3 & y&=-3n +63,
\end{align}
but different values of $a$ and $b$ are possible.
Can you see why different values of $a$ and $b$ are possible? Can you convince yourself that as $n$ varies any correct values of $a$ and $b$ will give all the solutions? You could also solve the equation $324 = 3x + 5y$ by drawing a graph and finding where it goes through points with whole number coordinates.
Since we know $0 < x < y$ we have the solutions:
| $x$ | $y$ | Total |
| $3$ | $63$ | $66$ |
| $8$ | $60$ | $68$ |
| $13$ | $57$ | $70$ |
| $18$ | $54$ | $72$ |
| $23$ | $51$ | $74$ |
| $28$ | $48$ | $76$ |
| $33$ | $45$ | $78$ |
| $38$ | $42$ | $80$ |
Well done Ewan! Remember that even though these solutions have come from looking at the algebra, they might not work. For instance neither DUM nor DEE can ever give away exactly $1/3$ of a pound! You could check the 'gifts' in each solution for $x$ and $y$ to make sure that this never happens.