## 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?

June 21st, 2013 at 3:28 pm

Starts on day 1 with 724 1/2 bananas

June 21st, 2013 at 4:29 pm

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

June 21st, 2013 at 7:18 pm

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

June 21st, 2013 at 7:35 pm

Attaboy Wiz. You doneded it.

June 24th, 2013 at 12:58 am

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.

June 24th, 2013 at 4:17 am

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