4 Stretch Your Thinking: Brian has some boxes of paper clips. Some boxes hold 10 clips and some boxes hold 100. He has some paper clips left over. He has three more boxes with 100 paper clips than he has boxes with 10 paper clips. He has two fewer paper clips left over than he has numbers of boxes with 100 paper clips. What number of paper clips could he have?

Relations are given between the number of 10-clip boxes, the number of 100-clip boxes, and the number left over. We assume there are at least one of each size box, and that there are fewer than 10 clips left over. (10 could be put into another box.)

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Let h represent the number of 100-clip boxes. Then h-3 is the number of 10-clip boxes, and h-2 is the number of clips left over. The total number of paperclips is ...

10(h -3) +100h +(h -2) = 111h -32

In addition, we require ...

h -3 > 0 and h-2 < 10

3 < h < 12 . . . . . . . rearranging these inequalities

So, Brian could have 111h -32 paper clips, where 3 < h < 12. That could be any of {412, 523, 634, 745, 856, 967, 1078, 1189} paper clips.