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2011-02-01T00:00:01
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<p>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?</p>
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<p>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.</p>
<p>Let $x$ be the three digit number at the start.</p>
<p>Let $y$ be the four digit number at the end of the phone
number.</p>
<p>The original phone number is $10000x + y$. The changed phone
number is $1000y + x$.</p>
<p>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.</p>
<p>We wrote the following program to test integer solutions:</p>
<pre>
<code>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
</code>
</pre>
<p>The answer is: 435 8717<br></br>
Elizabeth solved this equation $$998y-19999x = 1$$ using Euclid's
algorithm, as follows:</p>
<div class="math">\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}</div>
<p>Working backwards to get values for $x$ and $y$:</p>
<div class="math">\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}</div>
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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<mdoxml version="1.0"><p>The original number is $10000x + y$. What is the new number in terms of $x$ and $y$? Use the information to write down a Diophantine equation.</p>
<p> </p>
<p><a href="http://nrich.maths.org/public/viewer.php?obj_id=1357&part=index">(The article on Euclid's Algorithm might help with this problem.)</a></p></mdoxml>
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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?
Calculations and Numerical Methods
Euclid's algorithm
Algebra
Diophantine equations
Admin
Long problems
Numbers and the Number System
Place value