SOLUTION: I posted a problem here: https://www.algebra.com/tutors/students/your-answer.mpl?question=1204510 Ms.Ikelyn provided a similar problem, but I understand to little to find it u

Algebra ->  Customizable Word Problem Solvers  -> Misc -> SOLUTION: I posted a problem here: https://www.algebra.com/tutors/students/your-answer.mpl?question=1204510 Ms.Ikelyn provided a similar problem, but I understand to little to find it u      Log On

Ad: Over 600 Algebra Word Problems at edhelper.com


    

Question 1204514: I posted a problem here:
https://www.algebra.com/tutors/students/your-answer.mpl?question=1204510
Ms.Ikelyn provided a similar problem, but I understand to little to find it useful. Can someone walk me through everything on HER solution, or just solve the question.

Answer by greenestamps(13195) About Me  (Show Source):
You can put this solution on YOUR website!


The sum of the terms of an arithmetic sequence is

(number of terms) times (average of the terms)

If n is the average of the terms, then the sum of the terms of this sequence of 33075 consecutive integers is 33075n.

The problem requires that this sum be a perfect cube.

The prime factorization of 33075 is

33075=%283%5E3%29%285%5E2%29%287%5E2%29

For the sum 33075n to be a perfect cube, the exponents on all the prime factors of 33075n must be 3. So we need one more factor of 5 and one more of 7 to achieve that.

33075n=%283%5E3%29%285%5E2%29%287%5E2%29%285%2A7%29=%283%5E3%29%285%5E3%29%287%5E3%29

So the least positive average of the numbers that makes the sum a perfect cube is n = 5*7 = 35.

ANSWER: 35