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Difference between revisions of "2014 UNCO Math Contest II Problems/Problem 4"

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== Solution ==
 
== Solution ==
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(a) We can see that this is simply the 20th triangular number. Thus the answer is <math>\frac{20(21)}{2} = \boxed{210}</math>
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(b) From the previous problem, we see that the largest number written on the 20th slate is <math>210</math>. Since there are 20 numbers on the 20th slate, the smallest number is <math>210-20+1 = 191</math>. Since the numbers on the plate are consecutive, the answer is <math>20 \cdot \frac{191+210}{2} = \boxed{4010}</math>  (c) <math>\frac{n(n^2+1)}{2}</math>
  
 
== See also ==
 
== See also ==
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{{UNC Math Contest box|year=2014|n=II|num-b=3|num-a=5}}
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{{UNCO Math Contest box|year=2014|n=II|num-b=3|num-a=5}}
  
 
[[Category:Introductory Algebra Problems]]
 
[[Category:Introductory Algebra Problems]]

Latest revision as of 14:53, 28 October 2023

Problem

On the first slate, the Queen’s jurors write the number $1$. On the second slate they write the numbers $2$ and $3$. On the third slate the jurors write $4, 5$, and $6$, and so on, writing $n$ integers on the $n$th slate.

(a) What is the largest number they write on the $20$th slate?

(b) What is the sum of the numbers written on the $20$th slate?

(c) What is the sum of the numbers written on the $n$th slate?

Solution

(a) We can see that this is simply the 20th triangular number. Thus the answer is $\frac{20(21)}{2} = \boxed{210}$ (b) From the previous problem, we see that the largest number written on the 20th slate is $210$. Since there are 20 numbers on the 20th slate, the smallest number is $210-20+1 = 191$. Since the numbers on the plate are consecutive, the answer is $20 \cdot \frac{191+210}{2} = \boxed{4010}$ (c) $\frac{n(n^2+1)}{2}$

See also

2014 UNCO Math Contest II (Problems • Answer Key • Resources)
Preceded by
Problem 3 		Followed by
Problem 5
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10
All UNCO Math Contest Problems and Solutions
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