Prove that among any 54 integers 1 =< x1 < x2 < x3 <...< x54 =< 100, there is a pair with difference 12.
1 answer:
Use the pigeonhole principle.
Consider choosing
This selection guarantees that no two elements have a difference of 12. For example, in the first subset with
, the largest difference is 12 - 1 = 11. Then in the next subset with 
, the smallest difference between the least element of this subset with the previous subset is 25 - 12 = 13.
The last subset is problematic, however, because the last two elements of that subset are 48 + 53 = 101 and 48 + 54 = 102. We require that all the integers are no greater than 100, and that they are distinct, which means there must be two numbers
that belong to the set 
, and this will force at least one integer that makes up a pair with difference 12.
Consider choosing
This selection guarantees that no two elements have a difference of 12. For example, in the first subset with
The last subset is problematic, however, because the last two elements of that subset are 48 + 53 = 101 and 48 + 54 = 102. We require that all the integers are no greater than 100, and that they are distinct, which means there must be two numbers
You might be interested in
so we know she has sculptures and paintings, if she sold twice as many paintings as sculptures, that means that for every 2 paintings, she sold 1 sculpture, so the paintings and sculptures are on a 2:1 ratio.
we know she sold a total of 57, so we'll need to split 57 in a 2:1 ratio, we'll simply divide the whole amount of 57 by (2+1) and distribute accordingly.