The following problem results in a cubic, but is there a way to simplify it easier?:

The sum of the reciprocals of three consecutive integers is 47/60. What is the sum of these integers?

I ended up with $47{ x }^{ 3 }-180{ x }^{ 2 }-47x+60 =0$. However, unlike other methods where I've learned to "Let ${ x }^{ 2 }$ = k, $\therefore { x }^{ 4 }={ k }^{ 2 }$", ${x}^{3}$ is not a multiple of 2, so I cannot apply this same method. How should I proceed? (Other than just substitute, which the answers from this & this website did).


7 Answers 7


You can do just a little bit of trial and error. If the summands are $1/n, 1/(n + 1)$ and $1/(n + 2)$, then we ought to have

$$\frac 1 {n + 1} \approx \frac 1 3 \cdot \frac{47}{60} \approx \frac 1 4$$

because $1/(n + 1)$ is reasonably close to the average of the three reciprocals. And it turns out that

$$\frac 1 3 + \frac 1 4 + \frac 1 5 = \frac{47}{60}.$$


we have $$\frac{1}{n}+\frac{1}{n+1}+\frac{1}{n+2}=\frac{47}{60}$$ this equation is equivalent to $${\frac { \left( n-3 \right) \left( 47\,{n}^{2}+102\,n+40 \right) }{ 60\,n \left( n+1 \right) \left( n+2 \right) }} =0$$ can you finish now?

  • $\begingroup$ $\frac{47}{60}$, not $\frac{47}{6}$ $\endgroup$
    – user499203
    Nov 26, 2017 at 5:40
  • $\begingroup$ Ah thanks! I've got it. I guess trial and error is fine for this problem. $\endgroup$
    – DarkRunner
    Nov 26, 2017 at 5:46
  • 9
    $\begingroup$ @DarkRunner why use trial and error? Just solve the numerator equals to zero. Actually it is clear it has to be 3. Since (n-3) is zero for 3, and the second parenthesis is non-zero (positive) for any n>=0. $\endgroup$
    – lalala
    Nov 26, 2017 at 16:04

This answer is based upon symmetry and a guess.

We try to exploit symmetry and use the approach \begin{align*} \color{blue}{\frac{47}{60}}&=\frac{1}{n-1}+\frac{1}{n}+\frac{1}{n+1}\\ &=\frac{n(n+1)+n^2-1+(n-1)n}{n(n^2-1)}\\ &\color{blue}{=\frac{3n^2-1}{n(n^2-1)}} \end{align*}

Let's take a closer look at \begin{align*} \frac{3n^2-1}{n(n^2-1)}=\frac{47}{60} \end{align*}

Hoping for a simple proportion we equate the numerators and get \begin{align*} 3n^2-1=47\quad\Rightarrow\quad 3n^2=48\quad\Rightarrow\quad n\in\{4,-4\} \end{align*}

The denominator $n(n^2-1)$ evaluated at $n=4$ gives $4(16-1)=60$ and we get the solution $$\color{blue}{n=4}$$

Note: Observe, it was the symmetry which enabled us to make the lucky guess.


The greatest of the reciprocals must be at least $\frac 13 \cdot \frac {47}{60}=\frac {47}{180}\approx \frac 1{3.8}$, so the greatest has to be $\frac 11, \frac 12,$ or $\frac 13$. That is not many to try.


You can always solve the problem by factoring the cubic polynomial using standard methods for factoring polynomials. In this particular example notice that


  • 1
    $\begingroup$ If you notice that, then the problem becomes trivial. $\endgroup$
    – user193810
    Nov 26, 2017 at 11:24

Yet another method:



At this point, we can just take 60/i for i = 1,2,3,4,5,6:

60, 30, 20, 15, 12, 10

(We have to stop at 7 because 7 is not a factor of 60). We can then take consecutive sums of three numbers:

110, 65, 47, 37

Since 20+15+12 = 47, we can take 3,4,5 as our solution.

We could have also noted that the three term on the left side of the second equation are, on average, slightly less than 16, which would get us to the three terms closest to 16.

Also, don't forget about Eisenstein's criterion.


just take the LCM and form the LHS and compare the denominators of both the sides, that gives n=3 , then to verify put it in numerator.


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