1993 UNCO Math Contest II Problems/Problem 5

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 $28+24=\boxed{52}$

1993 UNCO Math Contest II (Problems • Answer Key • Resources)
Preceded by Problem 4	Followed by Problem 6
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1993 UNCO Math Contest II Problems/Problem 5

Contents

Problem

Solution

Solution 1

Solution 2

See also