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





Elizabeth solved this equation $$998y-19999x = 1$$ using Euclid's algorithm, as follows:
Working backwards to get values for $x$ and $y$:
Thus $y=8717$ and $x= 435$. The old telephone number is therefore 4358717. Checking, we have $$8717435 = 1 + 2\times 4358717.$$