A grocer likes to arrange his apples in nice rows. However, when he lays his apples in rows of 3 he has one left over. When he lays them in rows of 5 he also has one left over. The same thing happens when he arranges them in rows of 7 and 9. But when he arranges them in rows of 11, there are none left over. How many apples does the grocer have?

In order to arrange the apples evenly in rows of 11, our total number of apples should be a multiple of 11. For example, if we had 99 apples, we could make 9 rows of 11 with none left over. However, if we are to arrange the apples in rows of 3, 5, 7, and 9, and end up with one extra apple each time, our total should also be a multiple of 3, 5, 7, and 9, plus one more.

So, imagine for a moment that we take one apple away from our total and set it aside. The number of apples we have left would be a multiple of 3, 5, 7, and 9. This means we should focus our attention only on numbers which are multiples of all four of these. However, since any multiple of 9 is necessarily a multiple of 3, we don't need to concern ourselves with 3 as it is already accounted for.

So, what is the least common multiple of 5, 7, and 9? Well, these three numbers are pairwise coprime, which means that no two of them share any common factors greater than 1. In a case like this, the least common multiple will always be the product of all three numbers. So, our first possibility is 5 x 7 x 9 = 315. Now, if we return the single apple that we removed, we should have a multiple of 11. But in this case, we would have 316, which is not divisible by 11.

Moving on, the next number which is a multiple of 5, 7, and 9 would be 2 x 315 = 630. Adding the one remaining apple to this brings our total to 631, which is also not divisible by 11. So, we'll try the next possibility. 3 x 315 = 945 is the next largest multiple of 5, 7, and 9. Adding 1 to that gives us 946. This one turns out to be a multiple of 11, it is equal to 86.

So, the grocer has 946 apples.

-Nick Dale, Instructor