Jane Street quant interview problem | 105 pebbles

[@Quant_Prof]

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[@postselector]

For the second problem -- is it that the "chessboard sums" must be equal? By chessboard sum I mean that, if you 2-color the board with white and black squares, the sum of the white squares must equal the sum of the black squares.

[@Quant_Prof]

yes

[@vaibhavs2740]

You didn't mention the difficulty of this one, it seemed easy to me. Is the condition for problem 2 "sum of all = even " ?

[@Quant_Prof]

I would say the difficulty of the first one is medium-hard and the second one is medium. Your condition is necessary but not sufficient. For example, consider the 1x2 grid 2,4. You cant make both of them zero simultaneously.

[@YashDarak-z8w]

Can this be an alternate solution? The only way to reach a stack of 1 pebble is by dividing a stack of 2 pebbles to get two stacks of 1 pebble. So we can think of the step before the last as dividing 52 stacks of 2 into 104 stacks of 1. Where does the last stack of 1 come from? There is no way to form it Hence not possible. Had we needed an even number of 1 pebble stacks, this solution wouldn't hold but with an odd number of 1s, it isn't possible no matter what configuration we have at the start.

[@3tb5h6h]

Incorrect solution, imagine 5 pebbles a stqck of 2 and then 3 the two will split into 1 1 then one can be added to 3 stack making it 4 then 4 can be split to 2 2 then 1 1 1 1 You are expecting all to split together thats not the case By your theory you would need two sets of 2 2 and one individual which cannot come

[@YashDarak-z8w]

@3tb5h6h Ooooo yeah you're correct. Thank you!!