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Difference between revisions of "2020 AIME II Problems/Problem 10"

(Created page with "=Problem= Find the sum of all positive integers <math>n</math> such that when <math>1^3+2^3+3^3+\cdots +n^3</math> is divided by <math>n+5</math>, the remainder is <math>17</m...")
 
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=Problem=
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== Problem ==
 
Find the sum of all positive integers <math>n</math> such that when <math>1^3+2^3+3^3+\cdots +n^3</math> is divided by <math>n+5</math>, the remainder is <math>17</math>.
 
Find the sum of all positive integers <math>n</math> such that when <math>1^3+2^3+3^3+\cdots +n^3</math> is divided by <math>n+5</math>, the remainder is <math>17</math>.
 +
 +
==Solution 1 ==
 +
The formula for the sum of cubes, also known as Nicomachus's Theorem, is as follows:
 +
<cmath>1^3+2^3+3^3+\dots+k^3=(1+2+3+\dots+k)^2=\left(\frac{k(k+1)}{2}\right)^2</cmath>
 +
for any positive integer <math>k</math>.
 +
 +
So let's apply this to this problem.
 +
 +
Let <math>m=n+5</math>. Then we have
 +
<cmath>\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 \\
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\frac{400}{4}&\equiv 17 \mod m \\
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332 &\equiv 0 \mod m \\
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\end{align*}</cmath>
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So, <math>m\in\{83,166,332\}</math>. Testing the cases, only <math>332</math> fails. This leaves <math>78+161=\boxed{239}</math>.
 +
 +
<math>\LaTeX</math> and formatting adjustments and intermediate steps for clarification by Technodoggo.
 +
 +
==Solution 2 (Official MAA 1)==
 +
The sum of the cubes from 1 to <math>n</math> is
 +
<cmath>1^3+2^3+\cdots+n^3=\frac{n^2(n+1)^2}{4}.</cmath>For this to be equal to <math>(n+5)q+17</math> for some integer <math>q</math>, it must be that<cmath>n^2(n+1)^2=4(n+5)q+4\cdot 17,</cmath>so<cmath>n^2(n+1)^2 \equiv 4 \cdot 17= 68\hskip-.2cm \pmod{n+5}.</cmath>But <math>n^2(n+1)^2 \equiv (-5)^2(-4)^2 = 400 \pmod{n+5}.</math> Thus <math>n^2(n+1)^2</math> is congruent to both <math>68</math> and <math>400,</math> which implies that <math>n+5</math> divides <math>400-68 = 332=2^2 \cdot 83</math>. Because <math>n+5 > 17</math>, the only choices for <math>n+5</math> are <math>83, 166,</math> and <math>332.</math> Checking all three cases verifies that <math>n=78</math> and <math>n=161</math> work, but <math>n=327</math> does not. The requested sum is <math>78+161 = 239</math>.
 +
 +
==Solution 3 (Official MAA 2)==
 +
The sum of the cubes of the integers from <math>1</math> through <math>n</math> is<cmath>1^3+2^3+\cdots+n^3=\frac{n^2(n+1)^2}{4},</cmath>which, when divided by <math>n+5</math>, has quotient<cmath>Q=\frac14n^3 -\frac34n^2+4n-20 = \frac{n^2(n-3)}4+4n-20</cmath>with remainder <math>100.</math> If <math>n</math> is not congruent to <math>1\pmod4</math>, then <math>Q</math> is an integer, and<cmath>\frac{n^2(n+1)^2}{4} = (n+5)Q + 100 \equiv 17\pmod{n+5},</cmath>so <math>n+5</math> divides <math>100 - 17 =83</math>, and <math>n = 78</math>. If <math>n \equiv 1 \pmod4</math>, then <math>Q</math> is half of an integer, and letting <math>n = 4k+1</math> for some integer <math>k</math> gives<cmath>\frac{n^2(n+1)^2}{4} = 2(2k+3)Q + 100 \equiv 17\pmod{n+5}.</cmath>Thus <math>2k+3</math> divides <math>100-17 = 83</math>. It follows that <math>k=40</math>, and <math>n = 161</math>. The requested sum is <math>161 + 78 = 239</math>.
 +
 +
==Solution 4==
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Using the formula for  <math>\sum_{k=1}^n  k^3</math>,
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<cmath>1^3 + 2^3 + 3^3 + ... + n^3 = \frac{n^2(n+1)^2}{4}</cmath>
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Since <math>1^3 + 2^3 + 3^3 + ... + n^3</math> divided by <math>n + 5</math> has a remainder of <math>17</math>,
 +
<cmath>\frac{n^2(n+1)^2}{4} \equiv 17\pmod {n + 5}</cmath>
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Using the rules of modular arithmetic,
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<cmath>n^2(n+1)^2 \equiv 68\pmod {n + 5}</cmath><cmath>n^2(n+1)^2 - 68\equiv 0\pmod {n + 5}</cmath>
 +
Expanding the left hand side,
 +
<cmath>n^4 + 2 n^3 + n^2 - 68\equiv 0\pmod {n + 5}</cmath>
 +
This means that
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<math>n^4 + 2 n^3 + n^2 - 68</math> is divisible by <math>{n + 5}</math>.
 +
 +
<cmath>(n + 5) | (n^4 + 2 n^3 + n^2 - 68)</cmath>
 +
Dividing polynomials,
 +
<cmath>\frac{n^4 + 2 n^3 + n^2 - 68}{n + 5}</cmath>
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<cmath>= n^3 - 3 n^2 + 16n - 80 + \frac{332}{(n + 5)}</cmath>
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<math>(n + 5)</math>  <math>|</math>  <math>(n^4 + 2 n^3 + n^2 - 68)</math> <math>\iff</math> <math>\frac{332}{(n + 5)}</math> <math>\in</math> <math>\mathbb{Z}</math>
 +
<br><br>
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<math>\frac{332}{(n + 5)}</math> <math>\in</math>
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<math>\mathbb{Z}</math> <math>\iff</math> <math>(n + 5) = \pm 1, \pm 2, \pm 4, \pm 83, \pm 166, \pm 332</math>
 +
<br><br>
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Note that <math>n</math> <math>\in</math> <math>\mathbb{N}</math> and <math>n + 5 > 17</math> (because the remainder when dividing by <math>n + 5</math> is <math>17</math>, so <math>n + 5</math> must be greater than <math>17</math>), so all options <math>\leq 17</math> can be eliminated.
 +
<cmath>(n + 5) = 83, 166, 332</cmath>
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<cmath>n = 78, 161, 327</cmath>
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Checking all 3 cases, <math>n = 78</math> and <math>n = 161</math> work; <math>n = 327</math> fails.
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<br><br>
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Therefore, the answer is <math>78 + 161 = \boxed{239}</math>.
 +
<br><br>
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~ {TSun} ~
 +
 +
== Solution 5 (similar ideas to Solution 1, but faster) ==
 +
As before, we note that <math>(5+a)^3 + (n-a)^3 \equiv (5+a)^3 - (n+5 - (n-a))^3 \equiv 0 \pmod {n+5}.</math>
 +
Thus, we can pair up the terms from <math>5^3</math> to <math>n^3</math> and cancel them. We have to deal with two cases:
 +
 +
If <math>n</math> is even, then <math>5^3+6^3 + \cdots + n^3 \equiv 0 \pmod {n+5},</math> as there are an even number of terms and they pair and cancel. We thus get <math>1^2+2^3+3^3+4^3 = 100 \equiv 17 \pmod {n+5},</math> or <math>(n+5) | 83,</math> which yields <math>n=78.</math>
 +
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If <math>n</math> is odd, then <math>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}.</math>
 +
Letting <math>k = \frac{n+5}{2}</math> yields <math>k^2 + 83 \equiv 0 \pmod {2k}.</math> However, this means that <math>83</math> is divisible by <math>k,</math> so <math>k=1,83.</math>
 +
Plugging this back into <math>n</math> yields <math>n=2(83)-5 = 161</math> in the latter case.
 +
 +
Thus, the sum of all possible <math>n</math> is just <math>78+161 = \boxed{239}.</math>
 +
 +
- ccx09
 +
 +
== Video Solution by OmegaLearn ==
 +
https://artofproblemsolving.com/alcumus/problem
 +
 +
~ pi_is_3.14
 +
 +
==Video solution==
 +
https://www.youtube.com/watch?v=87Mp0cdUtCU
 +
~ North America Math Contest Go Go Go
 +
 +
==Video Solution==
 +
https://youtu.be/bz5N-jI2e0U?t=201
 +
 +
==See Also==
 +
{{AIME box|year=2020|n=II|num-b=9|num-a=11}}
 +
{{MAA Notice}}
 +
 +
[[Category:Intermediate Number Theory Problems]]

Latest revision as of 13:19, 21 October 2024

Problem

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

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 2 (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 3 (Official MAA 2)

The sum of the cubes of the integers from $1$ through $n$ is\[1^3+2^3+\cdots+n^3=\frac{n^2(n+1)^2}{4},\]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 4

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 5 (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

https://artofproblemsolving.com/alcumus/problem

~ pi_is_3.14

Video solution

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

Video Solution

https://youtu.be/bz5N-jI2e0U?t=201

See Also

2020 AIME II (Problems • Answer Key • Resources)
Preceded by
Problem 9 		Followed by
Problem 11
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All AIME Problems and Solutions

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