2022 AMC 10B Problems/Problem 14
Contents
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)
Let be 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.
Solution 3
The smallest sum of a number where is as we are using the smallest value of . Using this, we can say that if were an element of , then one of the sums (the smallest) would be . Thus must be the smallest element. So the largest amount of elements that could be in is the list of numbers from to as they all work. Because it is inclusive we have, .
~ Wiselion :)
Solution 4 (Pigeonhole v2)
We construct a possible subset with elements by including all odd integers from to , inclusive. . The sum of any elements is even, and thus cannot be an element of .
To show that cannot have more than elements, assume for sake of contradiction that . Let where and . Because the sums of any (not necessarily distinct) elements do not appear in , is not an element of for all . So, are all distinct integers. Let these integers be elements of the set . , and because , . But all elements of must be and , leaving only 26 possible values for the elements in . By the Pigeonhole Principle, the elements cannot be distinct, and we have a contradiction.
Thus, is the maximum possible size of .
~starwars101
Solution 5 (solution 2 with a deeper explanation. For if you felt stuck at the start)
The first element of subset must be . Nothing between and can possibly add up to , so that's your first number. Since you included , you cannot include and must instead include . Now that you have and , you cannot include or , so you include . Including rules out and . So the next numbers to include are and . By this point, it's evident that you're going by odds. But will this pattern will continue?
Since you started out with and then , you could only ever use those odd numbers to create even numbers. Odd plus odd is never odd, so the next smallest number can contain will always be the next smallest odd number.
So now you have the odds from to . You might realize quickly that that is elements, but if you don't, you can convert them to the evens to . By halving each term, you have integers to , so it's obvious now that the answer is
~mihikamishra
Video Solution
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
Video Solution by Interstigation
Video Solution by TheBeautyofMath
https://youtu.be/Mi2AxPhnRno?t=1056
~IceMatrix
See Also
2022 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 13 |
Followed by Problem 15 | |
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