Difference between revisions of "1983 AIME Problems/Problem 8"
Sevenoptimus (talk | contribs) (Cleaned up the solution (and deleted Solution 2, which, as far as I can tell, is simply wrong)) |
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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. | 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 2: Clarification of Solution 1 == | ||
== See Also == | == See Also == |
Revision as of 11:19, 14 August 2019
Problem
What is the largest -digit prime factor of the integer
?
Solution
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
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 |