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Two excellent solutions follow, one from Elizabeth Whitmore of Madras College, St Andrew's, which uses Euclid's algorithm and the other, which uses a computer program, from Serguey and Ilya from the International School of The Hague. First Serguey and Ilya's solution.

Let $x$ be the three digit number at the start.

Let $y$ be the four digit number at the end of the phone number.

The original phone number is $10000x + y$. The changed phone number is $1000y + x$.

The new number is one more than the old number doubled so $$20000x + 2y + 1 = 1000y + x$$ $$19999x + 1 = 998y.$$ There are an infinite number of solutions to this equation.

We wrote the following program to test integer solutions:

```
``````
Module1 - 1
Sub bbbbbbb ()
y = 2004
For x = 100 to 999
y = (19999 * x + 1) / 998
If y = Int(y) Then Debug.Print x; y
Next x
End Sub
```

The answer is: 435 8717

Elizabeth solved this equation $$998y-19999x = 1$$ using Euclid's
algorithm, as follows:

\begin{eqnarray} \\ {\bf 19,999} &=&
998\times 20 + 39\\ \mathbf{998} &=& \mathbf{39}\times 25 +
23\\ \mathbf{39} &=& \mathbf{23}\times 1 + 16\\ \mathbf{23}
&=& \mathbf{16}\times 1 + 7\\ \mathbf{16} &=&
\mathbf{7}\times 2 + 2\\ \mathbf{7} &=& \mathbf{2}\times 3
+ 1\\ \mathbf{2} &=& \mathbf{1}\times 2 + 0.
\end{eqnarray}

Working backwards to get values for $x$ and $y$:

\begin{eqnarray} 1 &=& {\bf 7}-3\times
{\bf 2}\\ &=& {\bf 7}-3\times ({\bf 16}-2\times {\bf 7})\\
&=& 7\times {\bf 7}-3\times {\bf 16}\\\ &=& 7\times
({\bf 23}-{\bf 16})-3\times {\bf 16}\\ &=& 7\times {\bf
23}-10\times {\bf 16}\\ &=& 7\times {\bf 23}-10\times ({\bf
39}-{\bf 23})\\ &=& 17\times {\bf 23}-10\times {\bf 39}\\
&=& 17\times ({\bf 998}-25\times {\bf 39})-10\times {\bf
39}\\ &=& 17\times {\bf 998}-435\times {\bf 39}\\
&=& 17\times {\bf 998}-435\times ({\bf 19999}-20\times {\bf
998})\\ &=& 8717\times {\bf 998}-435\times {\bf 19999}.
\end{eqnarray}

Thus $y=8717$ and $x= 435$. The old telephone number is therefore
4358717. Checking, we have $$8717435 = 1 + 2\times 4358717.$$