1993 UNCO Math Contest II Problems/Problem 5


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


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 (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