Problem:
How many ways are there to split the integers through into pairs such that in each pair, the greater number is at least times the lesser number?
Answer Choices:
A.
B.
C.
D.
E.
Solution:
If the lesser number in some pair is at least 8 , then the greater number has to be at least , which is impossible. This means that the lesser number in each pair must be between 1 and 7 , inclusive. Because there are exactly 7 pairs in the collection, it follows that the lesser numbers must be exactly the numbers 1 through 7 .
The 7 must be paired with the 14 . There are then 2 choices for the partner of 6 -being paired with either 12 or 13 . Now there are 3 choices for the partner of 5 , because exactly 2 of the numbers in have already been picked. Continuing in this fashion, there are 4 possible partners for the 4 , then 3 possible partners for the 3 , then 2 possible partners for the 2 , and finally 1 possible partner for the 1. Multiplying these numbers of choices together gives a final answer of .
The problems on this page are the property of the MAA's American Mathematics Competitions