Like solving brainteasers? 🧠 Give this one a crack! A clock face is broken into three pieces. Each piece has the same sum of numbers. Can you determine which numbers appear on each piece? For more brainteasers like this one, check out our sample trading interview questions at https://lnkd.in/gvWXPuZt
Austin Markwick
1mo
Piece 1: (11, 12, 1, 2), Piece 2: (3, 4, 9, 10), Piece 3: (5, 6, 7, 8). The sum of all numbers on a clock face is 78. 78/3 = 26, which is the sum of the numbers on each piece. Cutting the clock face using diagonal parallel lines, we get the three pieces above.
GradIQ
1mo
Got it? Or are you still piecing it together? 🤔
Pure Mathematics/CS @ ANU, Yale-NUS
1moIf this question is assuming the pieces are contiguous then there is no solution as the only contiguous piece containing 12 would have to be 11,12,1 and 2, since each of the pieces has to add to 1/3rd of the sum of the numbers 1-12, so each piece sums to 26. (You can verify why we can't start at 12 and go forwards). In this case however, starting from 3 and increasing there is no continuous subarray as 3+4+5+6+7=25. If we're assuming the pieces can be arbitrarily broken, then the solution is not unique, for example consider the segmentation [(11,12,1,2),(7,9,10),(3,4,5,6,8)] and [(11,12,3),(7,9,10),(1,2,4,5,6,8)] each are solutions. If we are considering them to be non contiguous but each piece is also of the same size (4 numbers), then a solution is [(11,12,1,2),(3,10,4,9),(5,8,6,7)], this is not unique as we can swap any two pairs of numbers in different pieces that sum to 13 can be swapped e.g. (3,10) and (5,8). There are however a finite number of solutions (clearly follows since there are finitely many permutations of 12 numbers ).