# Clock

### Clock

Divide the face of a circular clock into four parts with two lines which begin and end on the circle circumference, so that the sum of the numbers in the four parts are equal. Is it possible?
(there is also a shape showing a circle with the 12 numbers).

My thoughts: If we just add the 12 numbers, we get 78, for which the quotient of the division by 4 is not an integer.
Obviously we have to apply some of the following tricks:
First of all, it does not say "straight line", so it can also be any curve, with the only restriction that its starting and ending point must be on the circle.
We can split the 2-digit numbers with the line passing between them: In this way, 12 can be 1 + 2, 11 can be 1+1 and 10 can be 1+0 (not necessarily all together).
Also 6 and 9 are interchangeable (by rotation).

So for example, if we take the sum 1+...+8+6(instead of 9)+(1+0)+(1+1)+(1+2) we get 48, so each part must have a sum of 12, but I can't find any such arrangement!
Any ideas?

Alex.vollenga

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