You have a watch-face, numbered from 1 to 12 around a circle.

Can you with two cuts divide the face so you get the numbers in groups with the same sum as the others?

How do you do?
Is there more solutions than one?

Example: Cut in a cross and you get 1, 2, 3 in one group, 4, 5, 6 in another group etc. This is not a solution because 1+2+3=6 and 4+5+6=15 has not the same sum.

Originally posted by FabianFnas You have a watch-face, numbered from 1 to 12 around a circle.

Can you with two cuts divide the face so you get the numbers in groups with the same sum as the others?

How do you do?
Is there more solutions than one?

Example: Cut in a cross and you get 1, 2, 3 in one group, 4, 5, 6 in another group etc. This is not a solution because 1+2+3=6 and 4+5+6=15 has not the same sum.

Sum of all numbers is 78 which is not divisible by 4. We are therefore looking to divide the clockface into 3.

Unfortunately, the problem is how to do it with exactly TWO cuts.

I am wondering if there might be a second solution which involves cutting a two digit number into 2 separate numbers.. That might technically meet the qualifications.