Difference between revisions of "1995 IMO Problems/Problem 6"
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Let <math>p</math> be an odd prime number. How many <math>p</math>-element subsets <math>A</math> of <math>{1,2,\ldots,2p}</math> are there, the sum of whose elements is divisible by <math>p</math>? | Let <math>p</math> be an odd prime number. How many <math>p</math>-element subsets <math>A</math> of <math>{1,2,\ldots,2p}</math> are there, the sum of whose elements is divisible by <math>p</math>? | ||
− | ==Solution== | + | ==Solution 1 (Partition)== |
− | {{ | + | |
+ | Let <math>A(x,y)</math> be the generating function | ||
+ | |||
+ | <cmath>A(x,y) = (1+yx)(1+yx^2)\cdots(1+yx^{2p})</cmath> | ||
+ | |||
+ | We apply the roots of unity filter on <math>x</math> to get | ||
+ | |||
+ | <cmath>\frac{A(1,y)+A(w,y)+\cdots+A(w^{p-1},y)}{p} = \frac{(1+y)^{2p}+(p-1)(1+yw)\cdots(1+yw^{2p})}{p}</cmath> | ||
+ | |||
+ | We call this function on <math>y</math>, <math>B(y)</math>. Note that | ||
+ | |||
+ | <cmath>(1+w)(1+w^2)\cdots(1+w^{p}) = 2</cmath> | ||
+ | |||
+ | Then, we apply the roots of unity filter on <math>y</math> to get | ||
+ | |||
+ | \begin{align*} | ||
+ | \frac{B(1)+B(w)+B(w^2)+\cdots B(w^{p-1})}{p} &= \frac{p+p\binom{2p}{p}+p+2^{2}(p-1)(p)}{p^2} | ||
+ | \end{align*} | ||
+ | |||
+ | But, we need to subtract <math>2</math> because it counts the empty set and the set with size <math>2p</math>. This gives us | ||
+ | |||
+ | <cmath>\boxed{\frac{\dbinom{2p}{p}+2p-2}{p}}</cmath> | ||
+ | |||
+ | <math>\square</math> | ||
+ | |||
+ | Solution from this discussion: https://artofproblemsolving.com/community/c6t302107f6h15112_sum_of_whose_elements_is_divisible_by_p. | ||
==See Also== | ==See Also== | ||
{{IMO box|year=1995|num-b=5|after=Last Question}} | {{IMO box|year=1995|num-b=5|after=Last Question}} | ||
+ | |||
+ | * [[1995 IMO]] | ||
+ | * [[IMO Problems and Solutions]] | ||
+ | |||
+ | {{MAA Notice}} |
Latest revision as of 21:07, 18 January 2025
Problem
Let be an odd prime number. How many -element subsets of are there, the sum of whose elements is divisible by ?
Solution 1 (Partition)
Let be the generating function
We apply the roots of unity filter on to get
We call this function on , . Note that
Then, we apply the roots of unity filter on to get
\begin{align*} \frac{B(1)+B(w)+B(w^2)+\cdots B(w^{p-1})}{p} &= \frac{p+p\binom{2p}{p}+p+2^{2}(p-1)(p)}{p^2} \end{align*}
But, we need to subtract because it counts the empty set and the set with size . This gives us
Solution from this discussion: https://artofproblemsolving.com/community/c6t302107f6h15112_sum_of_whose_elements_is_divisible_by_p.
See Also
1995 IMO (Problems) • Resources | ||
Preceded by Problem 5 |
1 • 2 • 3 • 4 • 5 • 6 | Followed by Last Question |
All IMO Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.