Difference between revisions of "2024 AMC 10B Problems/Problem 12"
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~lprado | ~lprado | ||
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| + | ==Solution 2 (Rigorous)== | ||
| + | Let <math>n</math> be the total number of languages spoken by all the students, and let <math>k</math> be the number of languages that each student speaks. | ||
| + | Since all students speak the same number of languages, the condition given in the question can be modified as - | ||
| + | |||
| + | "If <math>N</math> is the <math>n</math>-element subset containing the union of the languages spoken by all students, each of the 100 students speaks a different (unique) <math>k</math>-element subset of <math>N</math> combination of languages." | ||
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| + | In more mathematical terms, this means <math>{n \choose k} \ge 100</math>. (Using [[Pigeonhole Principle|PHP]]) | ||
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| + | Because we need the least value of <math>n</math>, <math>k</math> must be the closest integer to <math>\frac{n}{2}</math> | ||
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| + | <math>\Rightarrow {n \choose \frac{n}{2}} \ge 100</math> for <math>n = 2m</math>, <math>{n \choose \frac{n-1}{2}} \ge 100</math> for <math>n=2m-1</math>, <math>m \in \mathbb{N}</math> | ||
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| + | In the context of this question, it will be fastest to just substitute <math>\normalsize n</math> to each of the option values to find the least value possible as <math>\boxed{\textbf{(A) } 9}</math> | ||
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| + | For a more rigorous proof, where options are not an option, one can manually count up from <math>n=1</math> as <math>100</math> is a small number. | ||
| + | |||
| + | ~laythe_enjoyer211 | ||
==See also== | ==See also== | ||
{{AMC10 box|year=2024|ab=B|num-b=11|num-a=13}} | {{AMC10 box|year=2024|ab=B|num-b=11|num-a=13}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 09:24, 14 November 2024
Problem
A group of 
students from different countries meet at a mathematics competition.
Each student speaks the same number of languages, and, for every pair of
students 
and 
, student speaks some language that student does not speak,
and student speaks some language that student does not speak. What is the
least possible total number of languages spoken by all the students?
Solution 1
Let's say we have some number of languages. Then each student will speak some amount of those languages, and no two people can have the same combination of languages or else the conditions will no longer be satisfied. Notice that 
. So each of the students can speak some 
of the 
languages. Thus, 
is our answer.
~lprado
Solution 2 (Rigorous)
Let 
be the total number of languages spoken by all the students, and let 
be the number of languages that each student speaks.
Since all students speak the same number of languages, the condition given in the question can be modified as -
"If 
is the -element subset containing the union of the languages spoken by all students, each of the 100 students speaks a different (unique) -element subset of combination of languages."
In more mathematical terms, this means 
. (Using PHP)
Because we need the least value of , must be the closest integer to 
for 
, 
for 
, 
In the context of this question, it will be fastest to just substitute 
to each of the option values to find the least value possible as 
For a more rigorous proof, where options are not an option, one can manually count up from 
as is a small number.
~laythe_enjoyer211
See also
| 2024 AMC 10B (Problems • Answer Key • Resources) | ||
| Preceded by Problem 11 |
Followed by Problem 13 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
| All AMC 10 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
