### Double Digit

Choose two digits and arrange them to make two double-digit numbers. Now add your double-digit numbers. Now add your single digit numbers. Divide your double-digit answer by your single-digit answer. Try lots of examples. What happens? Can you explain it?

### Pyramids

What are the missing numbers in the pyramids?

A little bit of algebra explains this 'magic'. Ask a friend to pick 3 consecutive numbers and to tell you a multiple of 3. Then ask them to add the four numbers and multiply by 67, and to tell you the last two digits of her answer. Now you can really amaze her by giving the whole answer and the three consecutive numbers used at the start.

# Weekly Problem 7 - 2010

##### Stage: 3 Short Challenge Level:
Since HCF$(m,n)=12$ we can write $m=12a$, $n=12b$ where $a$ and $b$ are coprime. The lowest common multiple of $m$ and $n$ must be $12ab$.

Since $12ab=4\times 3ab$, $3ab$ must be a square number, and since $a$, $b$ are coprime exactly one of them is a multiple of $3$. Without loss of generality suppose that $a$ is divisible by $3$ and write $a=3c$.

$3ab=9cb$ is a square number, so $cb$ is a square number, which implies that both $b$ and $c$ are square numbers, say $c=d^2$ and $b=e^2$. Therefore $m=12a=36d^2$ and $n=12e^2$.

Then $\frac{m}{4}=9d^2$ and $\frac{n}{3}=4e^2$ are square numbers, whereas $\frac{m}{3}=12d^2$, $\frac{n}{4}=3e^2$ and $mn=36\times 12d^2e^2$ are not.

This problem is taken from the UKMT Mathematical Challenges.

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