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Difference between revisions of "2018 UNCO Math Contest II Problems/Problem 9"

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elements in the set. For example, the three-element set <math>\{2, 48, 100\}</math> is not Grassilian, but the
 
elements in the set. For example, the three-element set <math>\{2, 48, 100\}</math> is not Grassilian, but the
 
six-element set <math>\{6, 10, 11, 20, 33, 39\}</math> is Grassilian. Let <math>G(n)</math> be the number of Grassilian subsets of <math>\{1, 2, 3, ..., n\}</math>. (By definition, the empty set is a subset of every set and is Grassilian.)
 
six-element set <math>\{6, 10, 11, 20, 33, 39\}</math> is Grassilian. Let <math>G(n)</math> be the number of Grassilian subsets of <math>\{1, 2, 3, ..., n\}</math>. (By definition, the empty set is a subset of every set and is Grassilian.)
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(a) Find <math>G(3)</math>, <math>G(4)</math>, and <math>G(5)</math>.
 
(a) Find <math>G(3)</math>, <math>G(4)</math>, and <math>G(5)</math>.
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(b) Find a recursion formula for <math>G(n + 1)</math>. That is, find a formula that expresses <math>G(n + 1)</math> in
 
(b) Find a recursion formula for <math>G(n + 1)</math>. That is, find a formula that expresses <math>G(n + 1)</math> in
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terms of <math>G(n), G(n 1),\;dots</math>
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terms of <math>G(n), G(n - 1),\dots</math>
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(c) Give an explanation that shows that the formula you give is correct.
 
(c) Give an explanation that shows that the formula you give is correct.
  
 
== Solution ==
 
== Solution ==
  
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a) <math>G(3) = 5, G(4) = 8, G(5) = 13</math>
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b) <math>G(n) = G(n - 1) + G(n - 2); G(0) = 1 \text{ and } G(1) = 2</math>
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c) <math>G(n + 1) = G(n) + G(n - 1)</math>.
  
 
== See also ==
 
== See also ==
 
{{UNCO Math Contest box|year=2018|n=II|num-b=8|num-a=10}}
 
{{UNCO Math Contest box|year=2018|n=II|num-b=8|num-a=10}}
  
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[[Category:]]
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[[Category:Intermediate Combinatorics Problems]]

Latest revision as of 20:02, 28 October 2023

Problem

Call a set of integers Grassilian if each of its elements is at least as large as the number of elements in the set. For example, the three-element set $\{2, 48, 100\}$ is not Grassilian, but the six-element set $\{6, 10, 11, 20, 33, 39\}$ is Grassilian. Let $G(n)$ be the number of Grassilian subsets of $\{1, 2, 3, ..., n\}$. (By definition, the empty set is a subset of every set and is Grassilian.)

(a) Find $G(3)$, $G(4)$, and $G(5)$.

(b) Find a recursion formula for $G(n + 1)$. That is, find a formula that expresses $G(n + 1)$ in terms of $G(n), G(n - 1),\dots$

(c) Give an explanation that shows that the formula you give is correct.

Solution

a) $G(3) = 5, G(4) = 8, G(5) = 13$

b) $G(n) = G(n - 1) + G(n - 2); G(0) = 1 \text{ and } G(1) = 2$

c) $G(n + 1) = G(n) + G(n - 1)$.

See also

2018 UNCO Math Contest II (Problems • Answer Key • Resources)
Preceded by
Problem 8 		Followed by
Problem 10
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10
All UNCO Math Contest Problems and Solutions
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