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a) A four digit number (in base 10) aabb is a perfect square. Discuss ways of systematically finding this number. (b) Prove that 11^{10}-1 is divisible by 100.

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What a Joke

Stage: 4 Challenge Level: Challenge Level:2 Challenge Level:2

First can I express my delight at the spreadsheet supplied by the Colyton Maths Challenge Group, which enabled them to enter values of $A$ and $H$ to find a solution. I think the cell H8 needed to be "=10*F3+D3" to work (and it does with the right substitution!) - but what a good idea. Well done. Like Angele (no school given), the Colyton group noticed that the problem might be easier with some rearrangement and making $JOKE$ the subject.

At this point it is clear that the maths group, like Andrei (Tudor Vianu College) used exhaustive methods, probably made easier by the spreadsheet.

Here is a alternative approach based on the solution offered by Lee (no school given).

I can rearrange the problem $$JOKE = \frac{AHHAAH}{HA}$$ Now $AHHAAH$ is made up of $AH00AH$ and $HA00 = 100 \times HA$ added together.

But $AH00AH$ is divisible by $AH$ So $AH00AH = 10001 \times AH$

So $AHHAAH = 10001 \times AH + HA \times 100$
$$JOKE = \frac{AHHAAH}{HA} = \frac{100 \times HA +AH \times(10001)}{HA}$$ The numerator has to be divisible by the denominator ($HA$) for this to give a four digit answer ($JOKE$). As $100 \times HA$ is divisble by $HA$ it only needs us to look at $10001 \times AH$.

But $10001$ is not prime, it is $73 \times 137$.

There is no combination of digits such that $HA$ divides $AH$ So $HA$ divides $10001$ giving an answer $73$

$$H=7$$ $$A=3$$ $$J=5$$ $$O=1$$ $$K=6$$ $$E=9$$