Difference between revisions of "1993 UNCO Math Contest II Problems/Problem 5"
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== Solution == | == Solution == | ||
− | + | The thirteenth integer is the average, which is <math>\frac{1000}{25}=40</math>. So, the largest integer is 12 larger, which is <math>40+12=\boxed{52}</math>, and the smallest integer is 12 less, which is <math>40-12=\boxed{28}</math>. | |
== See also == | == See also == |
Revision as of 16:47, 10 November 2017
Problem
A collection of consecutive positive integers adds to What are the smallest and largest integers in this collection?
Solution
The thirteenth integer is the average, which is . So, the largest integer is 12 larger, which is , and the smallest integer is 12 less, which is .
See also
1993 UNCO Math Contest II (Problems • Answer Key • Resources) | ||
Preceded by Problem 4 |
Followed by Problem 6 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 | ||
All UNCO Math Contest Problems and Solutions |