Arithmetic Sequence / Perfect Cube Question

[dagr8est]

The sum of 28 consecutive odd positive integers is a perfect cube. If p and q are the least and greatest of these integers, the average of the least possible value for p and the least possible value for q is between:
A) 80 and 90
B) 90 and 100
C) 100 and 110
D) 110 and 120
E) 120 and 130

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p is the smallest of the integers so p=t(1). q is the greatest of the integers so q=t(28).

t(n)=t(1)+d(n-1)
t(28)=p+2(28-1)
t(28)=p+54

Therefore, q=p+54.

S(n)=(n/2)[t(1)+t(n)]
S(28)=(28/2)(p+q)
S(28)=14(p+p+54)
S(28)=28p+756

This is where I'm stuck. The sum of the 28 consecutive odd positive integers is 28p+756, but I'm not sure how to determine what the lowest value of p that produces a perfect cube is. I've tried trial and error but with no success.

 

[Denis]

Why don't you start with a simple one and "see" what happens:
2^3 = 8 = 3 + 5
or:
4^3 = 64 = 13 + 15 + 17 +19

 

[Gene]

I don't, off hand, follow your steps. It looks like you might be using
the sum of the numbers from 1 to n = (n/2)(n+1) but I don't see how you are eliminating the even ones.
If it is correct you can use the fact that the sum of the odd numbers from 1 to q = q^2 so s(28) = q^2 - (p-1)^2 =
(p+54)^2 - (p^2-2p+1) =
p^2+54p+2916-p^2+2p-1 =
56p+2915
You would then have two equations in two unknowns and should be able to solve for p.

 

[dagr8est]

Why don't you start with a simple one and "see" what happens:
2^3 = 8 = 3 + 5
or:
4^3 = 64 = 13 + 15 + 17 +19

The smallest I found was 14^3=71+73+75...+123+125

That satisfies the terms of the question in that the sum of 28 consecutive odd integers is a perfect cube. Then:

(p+q)/2
=(71+125)/2
=98

B) Between 90 and 100

I was hoping there would be a better way than trial and error though.

 

[Gene]

I wrote

s(28) = q^2 - (p-1)^2 =
(p+54)^2 - (p^2-2p+1) =
p^2+54p+2916-p^2+2p-1 =
56p+2915

I got p & q mixed up with the pth & qth odd numbers. The number of q is (q+1)/2 and the number of the one below p is (p-1)/2 so the total fron p to q is
((q+1)/2)^2 - ((p-1)/2)^2 =
((q+1)^2-(p+1)^2)/4
q still = p+54 so we get
28p+756 = x^3 (the same as Dag had) or
2^2*7(p+27) = x^3
For the left side to be a cube
p+27 has to have factors of 2*7^2*y^3. The smallest y is 1 so
(p+27)=98
p=71 as Dag has found.
I put my mind into neutral when I thought we had the two equations or I would have thought of this earlier. I hope this gets rid of the trial and error.
---------------------
Gene

 

[soroban]

Hello, dagr8est!

The sum of 28 consecutive odd positive integers is a perfect cube.
If p and q are the least and greatest of these integers,
the average of the least possible value for p and the least possible value for q is between:
A) 80 and 90 . . B) 90 and 100 . . C) 100 and 110 . . D) 110 and 120 . . E) 120 and 130

- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -

. . . This is where I'm stuck.
The sum of the 28 consecutive odd positive integers is 28p+756,
but I'm not sure how to determine the lowest value of p that produces a perfect cube.
I've tried trial and error but with no success.

Great work!

We have: . 28(p + 27) .= . a³

. . . . or: . 2²·7(p + 27) .= .a³


We see that the left side has factors 2 and 7 . . . then so does a.
. . hence: a .= .2·7·b . . ---> . . a³ .= .2³·7³·b³

And we have: . 2²·7(p + 27) .= .2³·7³·b³ . . ---> . . p + 27 .= .98b³

The least value of p occurs when b = 1: . p = 71 . . . . There!