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Yellow streak

Posted by Chris on June 21, 2013 – 2:20 pm

As the last one was easy, here’s another already.

A banana seller sells half his bananas plus half a banana on the first day. He does the same the next day, and the next day, and so on, until he finally runs out of bananas at the end of the tenth day. How many bananas did he start with?


This post is under “Logic, Mathemagic” and has 6 respond so far.
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6 Responds so far- Add one»

  1. 1. JohnP24 Said:

    Starts on day 1 with 724 1/2 bananas

  2. 2. Chris Said:

    Nope. A clue is that he’s unlikely to actually have 1/2 bananas.

  3. 3. Wizard of Oz Said:

    On the last (tenth) day he has a single banana and sells half of this and gives a bonus half to end with zero.

    On the ninth day he has three bananas, sells 1 1/2 and gives a 1/2 to end up with the one for the last day.

    On the eighth day he has seven for 3 1/2 plus 1/2, and so on.

    So, following this sequence backwards, I make it that he starts with 1023 bananas.

    1023 – 511 – 255 – 127 – 63 – 31 – 15 – 7 – 3 – 1 – 0

  4. 4. Chris Said:

    Attaboy Wiz. You doneded it.

  5. 5. Werner Said:

    And the formula is b = 2^n – 1 where b is the number of banans he had n days ago if has has 0 today.

  6. 6. Chris Said:

    Hi Werner. That’s correct. I was hoping that someone would prove that formula.

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