2020 AIME II Problems/Problem 10


Find the sum of all positive integers $n$ such that when $1^3+2^3+3^3+\cdots +n^3$ is divided by $n+5$, the remainder is $17$.

Solution 1 (If you don't remember the formula for sum of cubes)

We first note that since the remainder is $17$ and we are dividing by $n+5$, $n+5$ must be greater than $17$, meaning that $n$ has to be at least $13$.

We then notice that we can pair the $5^3$ term with the $n^3$ term to factor it into $(n+5)(n^2-5n+25)$ using the sum of two cubes factorization ($a^3+b^3=(a+b)(a^2-ab+b^2)$), which is divisible by $n+5$. We can do the same for the $6^3$ term with the $(n-1)^3$ term, the $7^3$ term with the $(n-2)^3$, and so on, which are all divisible by $n+5$. However, when $n$ is odd, we will have a middle term that is not paired with any other terms, which is not necessarily divisible by $n+5$. Thus, we have two cases:

$\textbf{CASE 1: } n$ is even
If $n$ is even, all terms that are greater than $4^3$ pair, as there are an even number of terms that are greater than $4^3$. Therefore, all we need in order for the entire sequence to have a remainder of $17$ when divided by $n+5$ is $1^3+2^3+3^3+4^3$ to have a remainder of $17$ when divided by $n+5$.

Evaluating $1^3+2^3+3^3+4^3$ as $100$, all we need to be true is \[100\equiv 17\pmod {n+5},\] or that \[83\equiv 0\pmod {n+5}.\] Thus, $83$ will be divisible by $n+5$ where $n\geq 13$. As $83$ is prime, $n+5$ must be equal to either $1$ or $83$. If $n+5=1$, we have that $n=-4$, which is not greater than or equal to $13$, so that solution is extraneous.

If $n+5=83$, we have that $n=78$, which is $\geq 13$, so one of our solutions is $n=78$, and we are done with our first case.

$\textbf{CASE 2: } n$ is odd
If $n$ is odd, the only term that does not pair is the arithmetic mean of the numbers under the cube of the largest and smallest terms that would pair, or $\left(\frac{n+5}{2}\right)^3$. Therefore, as all other terms that are $\geq 5^3$ pair, the requirement that we have is \[1^3+2^3+3^3+4^3+\left(\frac{n+5}{2}\right)^3\equiv 17\pmod {n+5}.\] Calculating and simplifying, we have that \[83+\left(\frac{n+5}{2}\right)^3\equiv 0\pmod {n+5}.\] Now, we multiply both sides by $8$. However, since multiplication is not reversible in modular arithmetic, we need to check whether any solutions are extraneous after solving. The congruence that we now have is \[8\cdot 83+(n+5)^3\equiv 0\pmod {n+5}.\] As we know that $(n+5)^3$ is divisible by $n+5$, what we need now is \[8\cdot 83\equiv 0\pmod {n+5}.\] We now check each solution to see whether it works.

If $n+5=1, 2, 4, 8$, $n$ would be less than $13$, so none of these solutions work. If $n+5=83$, $n$ would be even, so that solution does not work for this case. Therefore, the only three solutions we need to check for this case are when $n+5=166$, $n+5=332$, or $n+5=664$. We plug these values into the congruence before we multiplied both sides by $8$ to check.

If $n+5=166$, we would need \[83+\left(\frac{166}{2}\right)^3\equiv 0\pmod {166}.\] Calculating and factoring out $83$, we have that \[83(1+83\cdot 83)\equiv 0\pmod {166}.\] As the right parenthesis is odd and $166=83\cdot 2$, we know that this solution works, so we have another solution: $n=166-5=161$.

If $n+5=332$, we would need \[83+\left(\frac{332}{2}\right)^3=83+166^3\equiv 0\pmod {322}.\] As the left hand side is odd, but all multiples of $322$ is even, this solution is therefore extraneous.

If $n+5=664$, we would need \[83+\left(\frac{664}{2}\right)^3=83+332^3\equiv 0\pmod {664}.\] Again, the left hand side is odd, and all multiples of $664$ are even, so this solution is extraneous.

Therefore, our final answer is $78+161=\boxed{239}$.

~ CoolCarsOnTheRun

Solution 2 (using the sum of cubes formula)

The formula for the sum of cubes, also known as Nicomachus's Theorem, is as follows: \[1^3+2^3+3^3+\dots+k^3=(1+2+3+\dots+k)^2=\left(\frac{k(k+1)}{2}\right)^2\] for any positive integer $k$.

So let's apply this to this problem.

Let $m=n+5$. Then we have \begin{align*} 1^3+2^3+3^3+\dots+(m-5)^3&\equiv 17 \mod m \\ \left(\frac{(m-5)(m-4)}{2}\right)^2&\equiv 17 \mod m \\ \left(\dfrac{m(m-9)+20}2\right)^2&\equiv 17\mod m \\ \left(\dfrac{20}2\right)^2&\equiv 17\mod m \\ \frac{400}{4}&\equiv 17 \mod m \\ 332 &\equiv 0 \mod m \\ \end{align*} So, $m\in\{83,166,332\}$. Testing the cases, only $332$ fails. This leaves $78+161=\boxed{239}$.

$\LaTeX$ and formatting adjustments and intermediate steps for clarification by Technodoggo.

Solution 3 (Official MAA 1)

The sum of the cubes from 1 to $n$ is \[1^3+2^3+\cdots+n^3=\frac{n^2(n+1)^2}{4}.\]For this to be equal to $(n+5)q+17$ for some integer $q$, it must be that\[n^2(n+1)^2=4(n+5)q+4\cdot 17,\]so\[n^2(n+1)^2 \equiv 4 \cdot 17= 68\hskip-.2cm \pmod{n+5}.\]But $n^2(n+1)^2 \equiv (-5)^2(-4)^2 = 400 \pmod{n+5}.$ Thus $n^2(n+1)^2$ is congruent to both $68$ and $400,$ which implies that $n+5$ divides $400-68 = 332=2^2 \cdot 83$. Because $n+5 > 17$, the only choices for $n+5$ are $83, 166,$ and $332.$ Checking all three cases verifies that $n=78$ and $n=161$ work, but $n=327$ does not. The requested sum is $78+161 = 239$.

Solution 4 (Official MAA 2)

The sum of the cubes of the integers from $1$ through $n$ is\[\frac{n^2(n+1)^2}{4},\]which, when divided by $n+5$, has quotient\[Q=\frac14n^3 -\frac34n^2+4n-20 = \frac{n^2(n-3)}4+4n-20\]with remainder $100.$ If $n$ is not congruent to $1\pmod4$, then $Q$ is an integer, and\[\frac{n^2(n+1)^2}{4} = (n+5)Q + 100 \equiv 17\pmod{n+5},\]so $n+5$ divides $100 - 17 =83$, and $n = 78$. If $n \equiv 1 \pmod4$, then $Q$ is half of an integer, and letting $n = 4k+1$ for some integer $k$ gives\[\frac{n^2(n+1)^2}{4} = 2(2k+3)Q + 100 \equiv 17\pmod{n+5}.\]Thus $2k+3$ divides $100-17 = 83$. It follows that $k=40$, and $n = 161$. The requested sum is $161 + 78 = 239$.

Solution 5

Using the formula for $\sum_{k=1}^n  k^3$, \[1^3 + 2^3 + 3^3 + ... + n^3 = \frac{n^2(n+1)^2}{4}\] Since $1^3 + 2^3 + 3^3 + ... + n^3$ divided by $n + 5$ has a remainder of $17$, \[\frac{n^2(n+1)^2}{4} \equiv 17\pmod {n + 5}\] Using the rules of modular arithmetic, \[n^2(n+1)^2 \equiv 68\pmod {n + 5}\]\[n^2(n+1)^2 - 68\equiv 0\pmod {n + 5}\] Expanding the left hand side, \[n^4 + 2 n^3 + n^2 - 68\equiv 0\pmod {n + 5}\] This means that $n^4 + 2 n^3 + n^2 - 68$ is divisible by ${n + 5}$.

\[(n + 5) | (n^4 + 2 n^3 + n^2 - 68)\] Dividing polynomials, \[\frac{n^4 + 2 n^3 + n^2 - 68}{n + 5}\] \[= n^3 - 3 n^2 + 16n - 80 + \frac{332}{(n + 5)}\] $(n + 5)$ $|$ $(n^4 + 2 n^3 + n^2 - 68)$ $\iff$ $\frac{332}{(n + 5)}$ $\in$ $\mathbb{Z}$

$\frac{332}{(n + 5)}$ $\in$ $\mathbb{Z}$ $\iff$ $(n + 5) = \pm 1, \pm 2, \pm 4, \pm 83, \pm 166, \pm 332$

Note that $n$ $\in$ $\mathbb{N}$ and $n + 5 > 17$ (because the remainder when dividing by $n + 5$ is $17$, so $n + 5$ must be greater than $17$), so all options $\leq 17$ can be eliminated. \[(n + 5) = 83, 166, 332\] \[n = 78, 161, 327\] Checking all 3 cases, $n = 78$ and $n = 161$ work; $n = 327$ fails.

Therefore, the answer is $78 + 161 = \boxed{239}$.

~ {TSun} ~

Solution 6 (similar ideas to Solution 1, but faster)

As before, we note that $(5+a)^3 + (n-a)^3 \equiv (5+a)^3 - (n+5 - (n-a))^3 \equiv 0 \pmod {n+5}.$ Thus, we can pair up the terms from $5^3$ to $n^3$ and cancel them. We have to deal with two cases:

If $n$ is even, then $5^3+6^3 + \cdots + n^3 \equiv 0 \pmod {n+5},$ as there are an even number of terms and they pair and cancel. We thus get $1^2+2^3+3^3+4^3 = 100 \equiv 17 \pmod {n+5},$ or $(n+5) | 83,$ which yields $n=78.$

If $n$ is odd, then $1^3+2^3+\cdots + n^3 \equiv 1^3+2^3+3^3+4^3+\left( \frac{n+5}{2} \right)^3 \equiv 17 \pmod {n+5}.$ Letting $k = \frac{n+5}{2}$ yields $k^2 + 83 \equiv 0 \pmod {2k}.$ However, this means that $83$ is divisible by $k,$ so $k=1,83.$ Plugging this back into $n$ yields $n=2(83)-5 = 161$ in the latter case.

Thus, the sum of all possible $n$ is just $78+161 = \boxed{239}.$

- ccx09

Video Solution by OmegaLearn


~ pi_is_3.14

Video solution

https://www.youtube.com/watch?v=87Mp0cdUtCU ~ North America Math Contest Go Go Go

Video Solution


See Also

2020 AIME II (ProblemsAnswer KeyResources)
Preceded by
Problem 9
Followed by
Problem 11
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png