$15 voucher giveaway. Optimum path.

Free giveaways treats. Deliver right to your house.

Just answer this question correctly. PM me your answer.

Pawpa has 3000 treats to deliver to a customer 1000 meters away. He brings chye with him. Pawpa’s hand is injured. So chye carries the treats, but chye can only carry  1000 at one go. And for every meter chye travels she’ll eat one treat up. Can pawpa ever deliver the treats? Why and why not. If yes, what’s the maximum number of treats he can finally deliver.

If you can formulate the rule to the optimal number, you will get $50.

4 responses to “$15 voucher giveaway. Optimum path.”

  1. The amount of treats left after each trip is (starting treats) – ceiling(treats/1000)*distance

    I assumed that each time Chye would carry 1000 (or as many as possible if the number of treats left is

    Then since the number of trips starts at 3 (3000/1000), we want to reduce the total number of trips that Chye makes. At treatos

    So I tried to find the closest distance where treatos

    When number of trips = 1, treatos

    She then makes one last trip to the end point, leaving her with 833 treatos 🙂


      • For the first step, was thinking something like this:
        Walks to 334m with 1000 treats (treats left = 666). Then walks back, gets another 1000 treats to 334m (left with another 666), walks back again and takes another 1000 (left with 666). Which brings her to a total of 1998 treats left
        Then she regroups this into 1000+998 (or 999+999, doesn’t matter). But let’s assume 1000+998
        Walk to 833m (distance = 449) with the first group of treats (left with 1000-449 = 551). Walk back to 334m and gets the 2nd group of 998 treats (left with 998-449 = 549 treats). Total treats = 551+549 = 1000
        Then walk the last 167m, leaving her with 833 treats
        By stop halfway you mean stop at 500m?


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