|
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):
Answer by ikleyn(52903)  (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  .
|
|
|
| |
