Difference between revisions of "2014 UNCO Math Contest II Problems/Problem 4"
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== Solution == | == Solution == | ||
| + | (a) We can see that this is simply the 20th triangular number. Thus the answer is <math>\frac{20(21)}{2} = \boxed{210}</math> | ||
| + | (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 == | ||
Latest revision as of 14:53, 28 October 2023
Problem
On the first slate, the Queen’s jurors write the number 
. On the second slate they write the
numbers 
and 
. On the third slate the jurors write 
, and 
, and so on, writing 
integers on
the th slate.
(a) What is the largest number they write on the 
th slate?
(b) What is the sum of the numbers written on the th slate?
(c) What is the sum of the numbers written on the th slate?
Solution
(a) We can see that this is simply the 20th triangular number. Thus the answer is 
(b) From the previous problem, we see that the largest number written on the 20th slate is 
. Since there are 20 numbers on the 20th slate, the smallest number is 
. Since the numbers on the plate are consecutive, the answer is 
(c) 
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 | ||
