Hi Annette,
This can be thought of as a guess and check problem. With the current amount of acorns, once the 7th chipmunk comes in and adds to the pile, 7 chipmunks can evenly divide their nuts. Thus, I would work backward from a multiple of 7.
The problem asks for the smallest amount of acorns that would work after the last chipmunk adds his acorn, so let’s start with the smallest multiple of 7 (7 acorns) and check it against the initial scenario:
“6 chipmunks try to divide their acorns evenly but have 5 acorns left.”
If we take the smallest multiple of 7 (7) and subtract one (before the last chipmunk adds his acorn), that brings us to the smallest possible amount of initial acorns. But it has to work with the problem's initial scenario. So let's check: If we divide this amount by 6, according to the problem, we’re supposed to have a remainder or 5. But if we divide 7 by 6, we have a remainder of 1, which does not work for the problem.
So let’s try the problem with the next smallest multiple of 7: 14.
14-1 (initial number or acorns) = 13. Divide 13 by 6, and get 2, remainder 1. We’re supposed to have 5 left over, so this does not work for the problem.
Let’s try the next smallest multiple of 7: 21.
21-1(initial number of acorns)=20. Divide 20 by 6, and get 3, remainder 2. We’re supposed to have 5 left over, so this does not work.
If we try this again, with each increasing multiples of 7, we eventually get the following:
42-1=41. Divide 41 by 6 and get 6, remainder 5. This is our initial amount of acorns. If we add one acorn back to this pile (the one that belongs to the 7th chipmunk), we will get 42 acorns, which is divisible by 7.
Thus, the smallest amount of acorns is 42. I hope this helps!
