Difference between revisions of "1983 AIME Problems/Problem 8"
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== Problem == | == Problem == | ||
− | What is the largest 2-digit prime factor of the integer <math>\ | + | What is the largest <math>2</math>-digit [[prime]] factor of the integer <math>n = {200\choose 100}</math>? |
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+ | ==Solution== | ||
+ | === Solution 1 === | ||
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+ | Expanding the [[combination|binomial coefficient]], we get <math>{200 \choose 100}=\frac{200!}{100!100!}</math>. Let the required prime be <math>p</math>; then <math>10 \le p < 100</math>. If <math>p > 50</math>, then the factor of <math>p</math> appears twice in the denominator. Thus, we need <math>p</math> to appear as a factor at least three times in the numerator, so <math>3p<200</math>. The largest such prime is <math>\boxed{061}</math>, which is our answer. | ||
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+ | === Solution 1.5, half (desperate) === | ||
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+ | The only way a single prime number will be left out after all the cancelation will have to satisfy the condition that out of all the multiples of the prime number we want to find, 2 of the multiples will have to be between 100 and 200, 61 is good for this solution(if you know your primes) because 61 times 2 and 6l times3 are both between 100-200 the reason for this is becauase if 2 copies of the prime we want to find are on numerater and one copy on the denominater, there will only be 1 copy of the prime we want to find when all the cancelation is made when reducing the origional fraction. | ||
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+ | <math>Michaellin16</math> | ||
− | == Solution == | + | === Solution 2: Clarification of Solution 1 === |
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− | + | We know that | |
+ | <cmath>{200\choose100}=\frac{200!}{100!100!}</cmath> | ||
+ | Since <math>p<100</math>, there is at least <math>1</math> factor of <math>p</math> in each of the <math>100!</math> in the denominator. Thus there must be at least <math>3</math> factors of <math>p</math> in the numerator <math>200!</math> for <math>p</math> to be a factor of <math>n=\frac{200!}{100!100!}</math>. (Note that here we assume the minimum because as <math>p</math> goes larger in value, the number of factors of <math>p</math> in a number decreases,) | ||
− | + | So basically, <math>p</math> is the largest prime number such that | |
+ | <cmath>\left \lfloor\frac{200}{p}\right \rfloor>3</cmath> | ||
+ | Since <math>p<\frac{200}{3}=66.66...</math>, the largest prime value for <math>p</math> is <math>p=\boxed{61}</math> | ||
− | + | ~ Nafer | |
− | + | ==Note== | |
− | + | Similar to 2023 MATHCOUNTS State Sprint #25 | |
− | == See | + | == See Also == |
− | + | {{AIME box|year=1983|num-b=7|num-a=9}} | |
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[[Category:Intermediate Combinatorics Problems]] | [[Category:Intermediate Combinatorics Problems]] | ||
+ | [[Category:Intermediate Number Theory Problems]] |
Latest revision as of 21:00, 26 June 2024
Contents
Problem
What is the largest -digit prime factor of the integer
?
Solution
Solution 1
Expanding the binomial coefficient, we get . Let the required prime be
; then
. If
, then the factor of
appears twice in the denominator. Thus, we need
to appear as a factor at least three times in the numerator, so
. The largest such prime is
, which is our answer.
Solution 1.5, half (desperate)
The only way a single prime number will be left out after all the cancelation will have to satisfy the condition that out of all the multiples of the prime number we want to find, 2 of the multiples will have to be between 100 and 200, 61 is good for this solution(if you know your primes) because 61 times 2 and 6l times3 are both between 100-200 the reason for this is becauase if 2 copies of the prime we want to find are on numerater and one copy on the denominater, there will only be 1 copy of the prime we want to find when all the cancelation is made when reducing the origional fraction.
Solution 2: Clarification of Solution 1
We know that
Since
, there is at least
factor of
in each of the
in the denominator. Thus there must be at least
factors of
in the numerator
for
to be a factor of
. (Note that here we assume the minimum because as
goes larger in value, the number of factors of
in a number decreases,)
So basically, is the largest prime number such that
Since
, the largest prime value for
is
~ Nafer
Note
Similar to 2023 MATHCOUNTS State Sprint #25
See Also
1983 AIME (Problems • Answer Key • Resources) | ||
Preceded by Problem 7 |
Followed by Problem 9 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |