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?
Solution
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 
See Also
| 2015 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 22 |
Followed by Problem 24 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
| All AJHSME/AMC 8 Problems and Solutions | ||
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