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## 'Double with 1-9' printed from http://nrich.maths.org/

The answer can be any one of:

$(6729,13458)$, $(6792,13584)$ , $(6927,13854)$, $(7269,14538)$,

$(7293,14586)$ , $(7329,14658)$ , $(7692,15384)$ , $(7923,15846)$,

$(7932,15864)$ , $(9267,18534)$ , $(9273,18546)$ , $(9327,18654)$

To try and reach one of these, you know that the first of the digits in the answer must be a $1$ that is carried. Any odd digit in the answer means the digit before in the original number must have been at least $5$.

Since we know where the $1$ must go, if $5$ was not in the answer, the number beneath it would either be $0$ or $1$, depending on whether there was a carry, but there is no $0$ and the $1$ has already been used.

After this you need to work systematically until you can find a solution.

*This problem is taken from the UKMT Mathematical Challenges.*