UNO High School Problem of the Week Competition
Problem 9
The algebraic sum of 16 consecutive, strictly positive integers is a perfect cube. Find the smallest value for the sum of these integers.
Solution
The numbers can be represented as (a)+(a+1)+(a+2)+...+(a+15)=16a+(15*8).
We can factor out an 8 from this equation so we have 8*(2a+15).
Since 8 is already a perfect cube, we need to find when 2a+15 is a perfect cube.
We know a cannot be 1 or 2, because 13 and 23 are less than 15 and a must be positive.
The equation 2a+15=33 has a positive integer solution of a=6, which means that the smallest
value in the sum of the integers is 6.
