Difference between revisions of "1983 AIME Problems/Problem 8"
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== Problem == | == Problem == | ||
What is the largest <math>2</math>-digit [[prime]] factor of the integer <math>n = {200\choose 100}</math>? | What is the largest <math>2</math>-digit [[prime]] factor of the integer <math>n = {200\choose 100}</math>? | ||
− | == Solution 1 == | + | ==Solution== |
− | Expanding the [[combination|binomial coefficient]], we get <math>{200 \choose 100}=\frac{200!}{100!100!}</math>. Let the 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 three times in the numerator, | + | === Solution 1 === |
+ | |||
+ | 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. | ||
+ | |||
+ | === Solution 2: Clarification of Solution 1 === | ||
+ | |||
+ | 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 | |
− | |||
== See Also == | == See Also == |
Revision as of 20:23, 14 January 2023
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 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
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 |