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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:
Working backwards to get values for $x$ and $y$:
The familiar Pythagorean 3-4-5 triple gives one solution to (x-1)^n + x^n = (x+1)^n so what about other solutions for x an integer and n= 2, 3, 4 or 5?