# Sunday Times Teaser 3014 – Family Business

*by Danny Roth*

#### Published Sunday June 28 2020 (link)

George and Martha run a company with their five daughters. The telephone extensions all have four positive unequal digits and strangely only four digits appear in the seven extensions:

\[\begin{array}{|l|r||}

\hline Andrea & abcd \\

\hline Bertha & acdb \\

\hline Caroline & bacd \\

\hline Dorothy & dabc \\

\hline Elizabeth & dbca \\

\hline George & cabd \\

\hline Martha & cdab \\

\hline \end{array}\]

They noticed the following:

Andrea’s and Bertha’s add up to Dorothy’s.

Bertha’s and Elizabeth’s add up to George’s.

Caroline’s and Dorothy’s add up to Martha’s.

What is Andrea’s extension?

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Here is a manual solution. The three constraints can be cast as three simultaneous equations:

\[1900a+91b+109c-989d=0\tag{1}\label{1}\]\[901a+91b-890c+1009d=0\tag{2}\label{2}\]\[190a+1009b-989c+901d=0\tag{3}\label{3}\]in the four positive digits \((a, b, c, d)\). Since we have three equations in four unknowns, we can’t fully solve them but we can use them to express any three of the variables in terms of the fourth.

Subtracting equation (\ref{2}) from equation (\ref{1}) and simplifying gives:\[a=2d-c\tag{4}\label{4}\]Subtracting 10 times equation (\ref{3}) from equation (\ref{1}) and simplifying gives\[b=c-d\tag{5}\label{5}\]Finally using equations (\ref{4}) and (\ref{5}) to eliminate \(a\) and \(b\) from equation (\ref{1}) gives:

\[5c=8d\tag{6}\label{6}\]

Since \(a\), \(b\), \(c\) and \(d\) are all single non-zero digits, equation (6) leads immediately to the solution \((c, d) = (8, 5)\) and equations (\ref{4}) and (\ref{5}) complete the full solution \((a, b, c, d)=(2, 3, 8, 5)\) with Andrea’s extension number as \(2385\).