BT.. Eat your heart out

If the last four digits of my phone number are placed in front of the remaining three you get one more than twice my number! What is it?

Age

16 to 18

Challenge level

Exploring and noticing Working systematically Conjecturing and generalising Visualising and representing Reasoning, convincing and proving

Being curious Being resourceful Being resilient Being collaborative

Problem

My phone number has seven digits: if the last four digits are placed in front of the remaining three you get one more than twice my number! What is the number?

Student Solutions

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 $$8717435 = 1 + 2\times 4358717.$$

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