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

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 $28+24=\boxed{52}$

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