# New Scientist Enigma 553 – Playing For Time

### by Chris Maslanka

#### From Issue #1706, 3rd March 1990

One disadvantage of doing recurrence relations in your head, Kugelbaum told himself, was that all to often you were still calculating away when the clock showed it was time to give a class. To buy himself a little more time he jotted down on the whiteboard:$\frac{A\widehat{X}B}{B\widehat{X}A}=\frac{AB}{BA}$
‘In this relation’, explained Kugelbaum to the class, ‘$$A$$, $$B$$ and $$X$$ are all non-zero digits and $$AB/BA$$ is a proper fraction in its lowest terms. The hat above the X in the numerator and that above the X in the denominator means that the digit $$X$$ is used but it must be repeatable an arbitrary number of times without affecting the validity of the equation. So the $$X$$ can be there once, twice, … 37 times … but, no matter how many times you use it, the equation must remain true. It is, or course, to be taken for granted that the number of $$X$$s interposed between the A and B must be the same for both the numerator and the denominator”.

So saying, the professor returned to his mental calculations while the class set about solving the problem.

How many distinct solutions does Kugelbaum’s equation have?

NB. A proper fraction has its numerator smaller than its denominator; if it’s in its lowest terms, numerator and denominator have no common factor other than 1.

One Comment
1. Let the centre $$X$$ components of the numerator and denominator of the left fraction have $$N$$ digits. The equation can then be recast as:

$\frac{(10B+A)(10^{N+1}A + 10(1+10+\dots+ 10^{N-1})X + B)}{(10A+B)(10^{N+1}B + 10(1+10+\dots+10^{N-1})X+A)}=1$

Expanding the numerator and denominator now gives:

$\frac{10^{N+1}(10AB+A^2)+10B^2+AB+(10/9)(10B+A)(10^N-1)X}{10^{N+1}(10BA+B^2)+10A^2 + BA+(10/9)(10A+B)(10^N-1)X}=1$

This equation means that we can equate the numerator and denominator and when we do so a number of terms cancel out to give:

$10(10^N-1)(A^2-B^2)=(10/9)(10^N-1)(9A-9B)X$and finally:$X = A + B$

This can easily be solved manually but here is a short program that shows that there are eight solutions: