# 99 Consecutive Positive Integers whose sum is a perfect cube?

What is the least possible value of the smallest of 99 consecutive positive integers whose sum is a perfect cube?

• What have you tried? What do you know about the sum of $99$ consecutive integers? If the first is $n$, what is the sum? – Ross Millikan Aug 19 '17 at 3:30
• Instead of 99, try solving the problem for only 9 consecutive numbers. – MJD Aug 19 '17 at 3:50

Hint 1: the sum of an odd number of consecutive integers is easiest described by the middle term. For example the sum of five consecutive integers where the middle term is $x$ is

$$(x-2)+(x-1)+x+(x+1)+(x+2)$$

$(x-2)+(x-1)+x+(x+1)+(x+2)=5x$. More generally, the sum of $n$ consecutive integers where $n$ is odd and $x$ is the middle term is $nx$

Hint 2: In a perfect cube, each prime must occur in the prime factorization a multiple of three number of times (zero is also a multiple of three)

$99=3^2\cdot 11^1$ is missing some factors to be a cube.

Let $\color{Blue}{n=3\cdot 11^2}\color{Red}{\cdot a}\color{Blue}{^3}$ for any arbitrary $\color{Red}{a}$. Only notice that $$\underbrace{ (n-49) + (n-48) + ... + (n-1) + \color{Blue}{n} + (n+1) + ... + (n+48) + (n+49)}_{\text{these are} \ \ 1+2\cdot 49 = 99 \ \ \text{consecutive numbers!}} \\ =99\color{Blue}{n}=99\cdot 3\cdot 11^2\cdot\color{Red}{a}^3=(33\color{Red}{a})^3.$$

Also one can prove that there are no other solutions!

• There are infinite solutions! After $35937=313+314+...$ there is $970299=9751+9752+...+9850$ and the others come from the $99n=3^2\times 11 \times n$ where $n$ is chosen to get a cube – Raffaele Aug 19 '17 at 14:06
• @Raffaele Yes you are right; for every $\color{Blue}{n=3\cdot 11^2}\color{Red}{\cdot a}\color{Blue}{^3}$; we get a solution ! – Davood Aug 19 '17 at 14:42

What an interesting question.

$(k+1) + (k+2) + (k+3)+......+(k+99) = n^3$

$99k + \sum_{j=1}^{99} j = n^3$

I suppose I shouldn't go any further and should let you figure it out from here... but now I am genuinely curious.

$99k + \frac {99*100}2 = n^3$

$99k + 4950 = n^3$

$99(k + 50) = n^3$.

So $99|n^3$ so $3^2|n^3$ and $3|n$ and $11|n$ so let $n = 33m$

$k + 50 = \frac {33^3}{99}m^3= 3*11^2*m^3$

the smallest possible value would be if $m = 1$ and $k = 3*11^2 -50=313$

And indeed $(313 + 1) + ..... + (313 + 99) = 35937 = 33^3$

So $314 + ..... + 412 = 99^3$ is the smallest such sum.

====

Argh. D'oh. $(j -49) + (j-48) + ...... + (j+48) + (j+49) = n^3$ is a sum of $99$ consecutive numbers (with $j - 48 = k+1$ if I were to compare to how I did it above).

The sum is $99*j = n^3$ so $j = 3*11^2$ is the smallest possible value and the first term is $3*11^2 - 49 = 314$.

• Read my hints above. $99^3=970299$ is not the smallest cube which is a multiple of $99=3^2\cdot 11^1$... $3^3\cdot 11^3=35937$ is. So, $j=99^2$ isn't the smallest possible value for the middle term, $j=3\cdot 11^2=363$ is. – JMoravitz Aug 19 '17 at 6:43
• D'oh, you are right. – fleablood Aug 19 '17 at 15:16