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##### Age 11 to 14Challenge Level

This is what happens, only twisted so that you know what's going on.

The friend picks three consecutive numbers, $x, x+1,$ and $x+2$ where $x \leq 58$.

He/she then picks a number $3n$, where $n \leq 33$ and tells you what it is.

Adding the 4 numbers gives $3x + 3 + 3n = 3(x + 1 + n)$.

Multiplying by 67 gives the number $201(x + 1 + n)$.

We know $x + 1 + n \leq 58 + 1 + 33 = 92$.

So the 2 digit number is 1 times $(x + 1 + n)$ and the rest is $200(x + 1 + n)$ which is 200 times the 2 digit number, so you double the 2 digit number to get the remaining digits.

You know what $n$ is, so subtract $n$ and 1 from the 2 digit number to get $x$.

From Ian Green, Age 13 Coopers Company and Coburn School, Upminster, Essex.