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Difference between revisions of "1993 UNCO Math Contest II Problems/Problem 5"

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== Solution ==
 
== Solution ==
To find the middle integer, we divide 1000 by 25 to get 40. We have found one number, so we know there are 24 more. We found the middle integer, so we know to find the largest and smallest we must add or subtract 24/2=12. Adding 12 gives us the largest is 52, and subtracting gives us the smallest is 28.
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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 $25$ consecutive positive integers adds to $1000.$ What are the smallest and largest integers in this collection?


Solution

The thirteenth integer is the average, which is $\frac{1000}{25}=40$. So, the largest integer is 12 larger, which is $40+12=\boxed{52}$, and the smallest integer is 12 less, which is $40-12=\boxed{28}$.

See also

1993 UNCO Math Contest II (ProblemsAnswer KeyResources)
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
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