2022 AMC 10B Problems/Problem 14
Problem
Suppose that 
is a subset of 
such that the sum of any two (not necessarily distinct)
elements of is never an element of . What is the maximum number of elements may contain?
Solution 1 (Pigeonhole Principle)
Denote by 
the largest number in .
We categorize numbers 
(except 
if is even) into 
groups, such that the 
th group contains two numbers and 
.
Recall that 
and the sum of two numbers in cannot be equal to , and the sum of numbers in each group above is equal to . Thus, each of the above groups can have at most one number in .
Therefore,
Next, we construct an instance of with 
.
Let 
.
Thus, this set is feasible.
Therefore, the most number of elements in is
.
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
Solution 2
We know that two odd numbers sum to an even number, so we can easily say that odd numbers 
can be included in the list, making for 
elements. But, how do we know we can't include even numbers for a higher element value? Well, to get a higher element value than , odd numbers as well as even numbers would have to be included in the list (since there are only 
even numbers from , and many of those even numbers are the sum of even numbers). However, for every even value we add to our odd list, we have to take away an odd number because there are either two odd numbers that sum to that even value, or that even value and another odd number will sum to an odd number later in the list. So, elements is the highest we can go.
Video Solution
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
See Also
| 2022 AMC 10B (Problems • Answer Key • Resources) | ||
| Preceded by Problem 13 |
Followed by Problem 15 | |
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| All AMC 10 Problems and Solutions | ||
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