Answer:
Let's assume that Brian must have fewer than 10 paperclips left over (otherwise he would have put them in a box of 10). Since there are only a few possible values for the number of paperclips Brian might have left over (any number from 1 to 9), we can just list them out.
If Brian has 1 paperclip left over, then he has 3 100s boxes (since there are two fewer leftovers than 100s boxes), but 0 10s boxes (since there are three more 100s boxes than 10s boxes). The problem told us he had some 10s boxes, so this scenario doesn't work.
If Brian has 2 paperclips left over, then he has 4 100s boxes (since there are two fewer leftovers than 100s boxes) and 1 10s box (since there are three more 100s boxes than 10s boxes).
If Brian has 3 paperclips left over, then he has 5 100s boxes and 2 10s boxes.
If Brian has 4 paperclips left over, then he has 6 100s boxes and 3 10s boxes.
If Brian has 5 paperclips left over, then he has 7 100s boxes and 4 10s boxes.
If Brian has 6 paperclips left over, then he has 8 100s boxes and 5 10s boxes.
If Brian has 7 paperclips left over, then he has 9 100s boxes and 6 10s boxes.
If Brian has 8 paperclips left over, then he has 10 100s boxes and 7 10s boxes.
If Brian has 9 paperclips left over, then he has 11 100s boxes and 8 10s boxes.
Each of these scenarios gives you a possible number of paperclips Brian might have. For example, with 2 leftovers, 4 100s boxes, and 1 10s box, Brian has a total of:
(4×100) + (1×10) + (2×1) = 400 + 10 + 2 = 412 paperclips
Step-by-step explanation:
Let's assume that Brian must have fewer than 10 paperclips left over (otherwise he would have put them in a box of 10). Since there are only a few possible values for the number of paperclips Brian might have left over (any number from 1 to 9), we can just list them out.
If Brian has 1 paperclip left over, then he has 3 100s boxes (since there are two fewer leftovers than 100s boxes), but 0 10s boxes (since there are three more 100s boxes than 10s boxes). The problem told us he had some 10s boxes, so this scenario doesn't work.
If Brian has 2 paperclips left over, then he has 4 100s boxes (since there are two fewer leftovers than 100s boxes) and 1 10s box (since there are three more 100s boxes than 10s boxes).
If Brian has 3 paperclips left over, then he has 5 100s boxes and 2 10s boxes.
If Brian has 4 paperclips left over, then he has 6 100s boxes and 3 10s boxes.
If Brian has 5 paperclips left over, then he has 7 100s boxes and 4 10s boxes.
If Brian has 6 paperclips left over, then he has 8 100s boxes and 5 10s boxes.
If Brian has 7 paperclips left over, then he has 9 100s boxes and 6 10s boxes.
If Brian has 8 paperclips left over, then he has 10 100s boxes and 7 10s boxes.
If Brian 2 = 412 paperclips
