2015 AMC 8 Problems/Problem 23
Tom has twelve slips of paper which he wants to put into five cups labeled , , , , . He wants the sum of the numbers on the slips in each cup to be an integer. Furthermore, he wants the five integers to be consecutive and increasing from to . The numbers on the papers are and . If a slip with goes into cup and a slip with goes into cup , then the slip with must go into what cup?
The numbers have a sum of , which averages to , which means will have values , respectively. Now it's process of elimination: Cup will have a sum of , so putting a slip in the cup will leave ; However, all of our slips are bigger than , so this is impossible. Cup has a sum of , but we are told that it already has a slip, leaving , which is too small for the slip. Cup is a little bit trickier, but still manageable. It must have a value of , so adding the slip leaves room for . This looks good at first, as we do have slips smaller than that, but upon closer inspection, we see that no slip fits exactly, and the smallest sum of 2 slips is , which is too big, so this case is also impossible. Cup has a sum of , but we are told it already has a slip, so we are left with , which is identical to the Cup C case, and thus also impossible. With all other choices removed, we are left with the answer: Cup
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