2
$\begingroup$

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

$\endgroup$
  • 1
    $\begingroup$ What have you tried? What do you know about the sum of $99$ consecutive integers? If the first is $n$, what is the sum? $\endgroup$ – Ross Millikan Aug 19 '17 at 3:30
  • $\begingroup$ Instead of 99, try solving the problem for only 9 consecutive numbers. $\endgroup$ – MJD Aug 19 '17 at 3:50
10
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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!

$\endgroup$
  • $\begingroup$ 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 $\endgroup$ – Raffaele Aug 19 '17 at 14:06
  • $\begingroup$ @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 ! $\endgroup$ – Davood Aug 19 '17 at 14:42
0
$\begingroup$

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 $3*11^2 - 49 = 314$.

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – JMoravitz Aug 19 '17 at 6:43
  • $\begingroup$ D'oh, you are right. $\endgroup$ – fleablood Aug 19 '17 at 15:16

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.