- #1
ehrenfest
- 2,020
- 1
Homework Statement
http://math.stanford.edu/~vakil/putnam07/07putnam1.pdf
I am working on number 6.
Consider the the ordered pairs (1,12),(2,13),...,(89,100). There are 89 of them. These are the pigeon holes.
There are 55 numbers between 1 to 100 and each of them flies to one or two pigeon holes. So split all of the pigeons not equal to 1 and 100 in half to get at least 108 pigeons. Then we have 108 pigeons flying to 89 holes. So there must be two numbers that differ by 11.
What is wrong with my logic?