SOLUTION: A series of 567 consecutive integers has a sum that is a perfect cube. Find the smallest possible positive sum for this series.

Algebra.Com
Question 1150758: A series of 567 consecutive integers has a sum that is a perfect cube. Find the smallest possible positive sum for this series.
Found 3 solutions by MathLover1, ikleyn, MathTherapy:
Answer by MathLover1(20850)   (Show Source): You can put this solution on YOUR website!
Let be the number, then the is , and the average is .
The sum is
Since the has to have and as a factor, the smallest value will be when

That "tops up" the factors of and to a multiple of , so that you get a .
The cube is .


Answer by ikleyn(52898)   (Show Source): You can put this solution on YOUR website!
.

Let  m+1, m+2, m+3, . . . . , n be a series of 567 consecutive integers with the sum of 567, with the first term (m+1),

where m is a positive integer, and the last term n,  n > m.


Then 

     -  = 

using the formula of the sum of the first positive integer numbers, with some integer positive k.


The formula (1) implies

     -  = 

     +  = 

    (n-m)*(n+m) + (n-m)         = 

    (n-m)*(n+m+1)               = .    (2)


Notice that  n-m = 567.  Thus the formula (2) becomes 

    567*(n+m+1)                 = .    (3)


The number 567 has the prime decomposition  567 = .


Therefore, in order for equation (3) be true with lowest possible value of (n+m-1), it should be

    n + m + 1 =  = 882.


Thus we have two equations

    n - m     = 567,      (4)

    n + m + 1 = 882.      (5)


By adding equations, you get  

    2n + 1 = 567 + 882 = 1449,

    2n = 1449-1 = 1448,

     n = 1448/2 = 724.


Then from equation (4),  m = 724 - 567 = 157.


Thus the sequence is

    158, 159, 160, . . . , 724.


Its sum is  =  =  =  = 250047.


ANSWER.  The smallest possible positive sum for this series is  = 250047.



Answer by MathTherapy(10557)   (Show Source): You can put this solution on YOUR website!
A series of 567 consecutive integers has a sum that is a perfect cube. Find the smallest possible positive sum for this series. 
Yet, ANOTHER method!!

Sum of an A.P.:	

                 ------ Substituting 567 for n, and 1 for d

                 ======>  ======>  =====> 

		 ------- Substituting PRIME FACTORS

                 ------ Factoring out GCF, 34 * 7

From above, it can be seen that a PERFECT CUBE of base 3 would be 36, so ANOTHER 32 is needed (to be MULTIPLIED), and a PERFECT CUBE of base 7 would be 73,

and so, ANOTHER 72 is needed (to be MULTIPLIED) also.

Therefore, for the SMALLEST CUBE, we need to have: 36 * 73, or , which is actually . 

RELATED QUESTIONS


A series of 567 consecutive integers has a sum that is a perfect cube. Find the smallest... (answered by Alan3354)

A series of 384 consecutive odd integers has a sum that is a perfect fourth power of a... (answered by ikleyn,greenestamps)

A series of 384 consecutive odd integers has a sum that is a perfect fourth power of a... (answered by MathLover1,ikleyn,greenestamps)

A series of 10584 consecutive integers has a sum that is a perfect cube. What is the... (answered by greenestamps)

A series of 33075 consecutive integers has a sum that is a perfect cube. What is the... (answered by ikleyn)

A series of 384 consecutive odd integers has a sum that is a perfect fourth power of a... (answered by Edwin McCravy)

The sum of the 28 consecutive odd, positive integers is a perfect cube. Find the smallest  (answered by Alan3354)

An arithmetic series of 30 375 terms has a sum that is a perfect fifth power. What is the  (answered by Edwin McCravy,ikleyn)

The sum of 18 consecutive odd integers is a perfect fifth power of n. If x is the... (answered by greenestamps)