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

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A collection of <math>25</math> consecutive positive integers adds to <math>1000.</math> What are the smallest and largest integers in this collection?
 
A collection of <math>25</math> consecutive positive integers adds to <math>1000.</math> What are the smallest and largest integers in this collection?
  
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== Solution ==
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=== Solution 1 ===
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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>.
  
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== Solution ==
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=== Solution 2 ===
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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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By the summation formula, the sum of 25 consecutive numbers (where <math>x</math> is the smallest number in the list) is
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<cmath>\frac{25(2x+24)}{2}</cmath>
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<cmath>25(x+12)</cmath>
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Letting the value of the equation be <math>1000</math>, we have
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<cmath>25(x+12)=1000</cmath>
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<cmath>x+12=40</cmath>
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<cmath>x=28</cmath>
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Thus the smallest value of the list is <math>\boxed{28}</math>, and the largest is <math>28+24=\boxed{52}</math>
  
 
== See also ==
 
== See also ==

Latest revision as of 00:05, 20 January 2023

Problem

A collection of $25$ consecutive positive integers adds to $1000.$ What are the smallest and largest integers in this collection?

Solution

Solution 1

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 $40-12=\boxed{28}$.

Solution 2

By the summation formula, the sum of 25 consecutive numbers (where $x$ is the smallest number in the list) is \[\frac{25(2x+24)}{2}\] \[25(x+12)\] Letting the value of the equation be $1000$, we have \[25(x+12)=1000\] \[x+12=40\] \[x=28\] Thus the smallest value of the list is $\boxed{28}$, and the largest is $28+24=\boxed{52}$

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
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