Difference between revisions of "1983 AIME Problems/Problem 8"
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We know that | We know that | ||
<cmath>{200\choose100}=\frac{200!}{100!100!}</cmath> | <cmath>{200\choose100}=\frac{200!}{100!100!}</cmath> | ||
+ | Since <math>p<100</math>, there is a 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>. | ||
== See Also == | == See Also == |
Revision as of 12:17, 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
We know that
Since
, there is a 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
.
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 |