The first one who came up with the correct number of apples was *** Diapason *** who was kind enought to mail me the answer, in order not to spoil the fun for the rest of you. Well done!
Yes, you're right. The minimal solution is 946, and can be calculated with the Chinese Remainder Theorem as given by Dejection.
There are more solutions, of course, I didn't explicitly ask for the minimum solution, sorry for that.
When I put my apples in 2 piles of the same number of apples in each pile, I got 1 apple left.
When I put my apples in 3 piles of the same number of apples in each pile, I got 2 apples left.
When I put my apples in 4 piles of the same number of apples in each pile, I got 3 apples left.
When I put my apples in 5 piles of the same number of apples in each pile, I got 4 apples left.
When I put my apples in 6 piles of the same number of apples in each pile, I got 5 apples left.
When I put my apples in 7 piles of the same number of apples in each pile, I got 6 apples left.
When I put my apples in 8 piles of the same number of apples in each pile, I got 7 apples left.
When I put my apples in 9 piles of the same number of apples in each pile, I got 8 apples left.
When I put my apples in 10 piles of the same number of apples in each pile, I got 9 apples left.
Originally posted by FabianFnas When I sell my apples in the market I am very much into symmetry. I always put them in piles of the same amount in order to show them nicely.
So today I put them in 3 piles of the same number of apples in each pile. But I got one apple left.
I tried 5 piles, but the same happened, one apple too many to be able to divide them evenly.
I tried 7 piles, ...[text shortened]... same number of apples in them. No apple too many.
How many apples did I bring to the market?
If you had counted the apples before you left your house you wouldn't have to ask such a silly question.