Greengrocer C. Carrot wants to expose his oranges neatly for sale. Doing this he discovers that one orange is left over when he places them in groups of three. The same happens if he tries to place them in groups of 5, 7, or 9 oranges. Only when he makes groups of 11 oranges, it fits exactly.

The Question: How many oranges does the greengrocer have at least?

Answer: Assume the number of oranges is A. Then A-1 is divisible by 3, 5, 7 and 9. So, A-1 is a multiple of 5×7×9 = 315 (note: 9 is also a multiple of 3, so 3 must not be included!). We are looking for a value of N for which holds that 315×N + 1 is divisible by 11. After some trying it turns out that the smallest N for which this holds is N = 3. This means that the greengrocer has at least 946 oranges.

Note that for N = 14, 25, 36, etc. (so each time 11 more) it also holds that 315×N + 1 is divisible by 11.

## Saturday, July 26, 2008

### Odd Oranges

Posted by Covert Bay at 2:49 AM

Labels: Riddles/Puzzles

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