The sum of the cube of three consecutive numbers is 104 squared. What are these numbers?

Steps to Solve

1. Define the consecutive numbers

Let the three consecutive numbers be $n-1$, $n$, and $n+1$. This makes the algebra easier.

2. Set up the equation

The sum of their cubes is equal to $104^2$, so we have: $$ (n-1)^3 + n^3 + (n+1)^3 = 104^2 $$ $$ (n-1)^3 + n^3 + (n+1)^3 = 10816 $$

3. Expand the cubic terms

Expand $(n-1)^3$ and $(n+1)^3$: $$ (n-1)^3 = n^3 - 3n^2 + 3n - 1 $$ $$ (n+1)^3 = n^3 + 3n^2 + 3n + 1 $$

4. Substitute and simplify the equation

Substitute these expansions into the equation: $$ (n^3 - 3n^2 + 3n - 1) + n^3 + (n^3 + 3n^2 + 3n + 1) = 10816 $$ Simplify by combining like terms: $$ 3n^3 + 6n = 10816 $$

5. Further simplification

Divide the entire equation by 3: $$ n^3 + 2n = \frac{10816}{3} $$ $$ n^3 + 2n = 3605.333... $$ Since we are looking for consecutive integers, it is likely there was a typo in the original question, and it was intended that the result of the sum of cubes be an integer as well. Let's round to the nearest whole number with the knowledge that this may be an estimation. Since it is closest to 3605, we will use that number. $$ n^3 + 2n = 3605$$

6. Solve for n

We are looking for an integer solution, we can try to guess and check values of $n$ close to $\sqrt[3]{3605} \approx 15.33$. Try $n = 15$: $$ 15^3 + 2(15) = 3375 + 30 = 3405 $$ Try $n = 16$: $$ 16^3 + 2(16) = 4096 + 32 = 4128 $$ Since 3605 is closer to 3405 than 4128, we can estimate that $n$ is close to 15, so our consecutive integers are close to 14, 15, and 16. We know there was a likely typo or rounding involved, so we now adjust backwards.

Going back to our original equation $3n^3 + 6n = 10816 $, dividing by 3 gave us recurring decimals, so let's instead divide by 6 to look at something simpler. We get: $(\frac{10816}{6}) = 1802.666...$ So $ \frac{n^3}{2} + n = 1802.666... $ $\frac{n^3}{2}$ is close to 1800, so once again our cube root is going to be close to 15. Let us check the numbers around 10816 to see if one of them is divisible by 3. 10815/3 = 3605, 10818/3 = 3606, 10817/3 = 3605.666... So perhaps a reasonable edit would be that $104^2 = 10816$ was intended to be $10815$, or $3605*3$.

7. Re-solve for n with the edited constraints $$ (n-1)^3 + n^3 + (n+1)^3 = 10815 $$ $$ (n^3 - 3n^2 + 3n - 1) + n^3 + (n^3 + 3n^2 + 3n + 1) = 10815 $$ $$ 3n^3 + 6n = 10815 $$ $$ n^3 + 2n = 3605 $$ Let's check $n = 15$: $$ 15^3 + 2(15) = 3375 + 30 = 3405 $$ It looks like we need to find a number $x$ that when added to 3405 becomes 3605. $ x = 3605-3405 = 200 $. $x$ needs to be added to one of the expressions.

Let us try $n = 14$: $$ 14^3 + 2(14) = 2744 + 28 = 2772 $$

8. Solve by approximation Instead of consecutive integers, let us assume consecutive real numbers. $$ n^3 + 2n = 3605$$ $$ 15.333^3 + 2(15.333) = 3616 + 30.666 = 3646.666 $$ $$ 15.2^3 + 2(15.2) = 3511.808 + 30.4 = 3542.208 $$ $$ 15.25^3 + 2(15.25) = 3558.89 + 30.5 = 3589.39 $$ $$ 15.26^3 + 2(15.26) = 3568.2 + 30.52 = 3598.72$$ $$ 15.27^3 + 2(15.27) = 3577.5 + 30.54 = 3608.04$$ So although we could continue approximating, we should go back and check for errors, since according to the question we are looking for integers. Alternatively, we should ask for clarification as the problem seems to implicitly require integer solutions, which do not seem to exist.

9. Solving the Edited Problem, again

Since dividing the original equation gives $ n^3 + 2n = 3605 $, then by inspection, $n=15$ is too small since it gives $3405$, and $n=16$ is too big since it gives $4128$, then the three real numbers $n-1, n, n+1$ are not integers, and also do not sum to 104^2 = 10816 as stated. If the goal of the question was $10815$ instead, then no integers solve this equation. $$ (n-1)^3 + n^3 + (n+1)^3 = 3n^3 + 6n$$ If the sum of cubes is $10815$, then we still require that $n$ be an integer.

10. Conclusion

We believe there has been a typo somewhere in the problem, as with the edited constraints, we are unable to find an integer solution. Further clarification is required. For now it's best to leave the answer at 15.27.