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| A set of numbers is called <imath>sum</imath>-<imath>free</imath> if whenever <imath>x</imath> and <imath>y</imath> are (not necessarily distinct) elements of the set, <imath>x+y</imath> is not an element of the set. For example, <imath>\{1,4,6\}</imath> and the empty set are sum-free, but <imath>\{2,4,5\}</imath> is not. What is the greatest possible number of elements in a sum-free subset of <imath>\{1,2,3,...,20\}</imath>?
| | #redirect [[2025 AMC 12A Problems/Problem 15]] |
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| <imath>\textbf{(A) } 8 \qquad\textbf{(B) } 9 \qquad\textbf{(C) } 10 \qquad\textbf{(D) } 11 \qquad\textbf{(E) } 12</imath>
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| ==Solution 1==
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| Let our subset be <imath>\{11,12,13,...,20\}.</imath> If we add any one element from the set <imath>\{1,2,3,...,10\},</imath> we will have to remove at least one element from our current subset. Hence, the size of our set cannot exceed <imath>\boxed{\text{(C) }10}.</imath>
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| ~Tacos_are_yummy_1
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| ==Solution 2==
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| Let our subset be <imath>\{1,3,5,...,19\}.</imath> Since odd numbers + odd numbers will always sum to an even number, this subset holds true. Leo, the addition of any even number will result in a violation of the rule, so the maximum number of elements is <imath>\boxed{\text{(C) }10}.</imath>
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| ~Kevin Wang
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| ==Solution 3 (Those who know)==
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| I'm kinda surprised that this question is just the copy-and-paste version of 2022 10B Problem 14. This problem is easier yet it's at problem 21. Nice problem quality you got there, huh, just adding a fancy definition, and yay, you got a brand new problem!
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| https://artofproblemsolving.com/wiki/index.php?title=2022_AMC_10B_Problems/Problem_14
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| ~metrixgo
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| I litterly remember the problem from 2022. If anything, this one's easier. Not complaining though.
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| Its literally just the number of elements if you only put the odd numbers in like what this should’ve been problem 10 not that semicircle problem
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| ~[[User:grogg007|grogg007]]
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| frfr I thought so too
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| ~BOTNATE
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| == Video Solution (In 1 Min) ==
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| https://youtu.be/V_zh78Ae8xw?si=D8dEsX4ST3JORj6x ~ Pi Academy
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| ==Video Solution==
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| https://youtu.be/gWSZeCKrOfU
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| ~MK
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| ==See also==
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| {{AMC10 box|year=2025|ab=A|num-b=20|num-a=22}}
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| {{AMC12 box|year=2025|ab=A|num-b=17|num-a=19}}
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| {{MAA Notice}}
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