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

See also

2014 UNCO Math Contest II (ProblemsAnswer KeyResources)
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