Difference between revisions of "2020 CIME I Problems/Problem 7"
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==Solution== | ==Solution== | ||
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| − | + | Notice that the product of first 2n+1 odd numbers = (2n+2)! / ((2^n+2)*(n+1)!) ; substituting this for f(n) gives the answer | |
==See also== | ==See also== | ||
{{CIME box|year=2020|n=I|num-b=6|num-a=8}} | {{CIME box|year=2020|n=I|num-b=6|num-a=8}} | ||
Revision as of 23:37, 18 December 2020
Problem 7
For every positive integer 
, define ![\[f(n)=\frac{n}{1 \cdot 3 \cdot 5 \cdots (2n+1)}.\]](http://latex.artofproblemsolving.com/4/e/6/4e66db1ce558cf1817f1579286937d4ed44e956b.png)
Suppose that the sum 
can be expressed as 
for relatively prime integers 
and 
. Find the remainder when is divided by 
.
Solution
This problem needs a solution. If you have a solution for it, please help us out by adding it. Notice that the product of first 2n+1 odd numbers = (2n+2)! / ((2^n+2)*(n+1)!) ; substituting this for f(n) gives the answer
See also
| 2020 CIME I (Problems • Answer Key • Resources) | ||
| Preceded by Problem 6 |
Followed by Problem 8 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
| All CIME Problems and Solutions | ||
The problems on this page are copyrighted by the MAC's Christmas Mathematics Competitions. 
