2014 UNCO Math Contest II Problems/Problem 4

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) $\frac{n(n^2+1)}{2}$

2014 UNCO Math Contest II (Problems • Answer Key • Resources)
Preceded by Problem 3	Followed by Problem 5
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2014 UNCO Math Contest II Problems/Problem 4

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