Difference between revisions of "2020 AIME II Problems/Problem 10"
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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> | 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: | 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: | ||
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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 these 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> | 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 these 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> | 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> | 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> |
Revision as of 00:45, 7 March 2021
Contents
Problem
Find the sum of all positive integers such that when is divided by , the remainder is .
Solution 1 (If you don't remember the formula for sum of cubes)
We first note that since the remainder is and we are dividing by , must be greater than , meaning that has to be at least .
We then notice that we can pair the term with the term to factor it into using the sum of two cubes formula (), which is divisible by . We can do the same for the term with the term, the term with the , and so on, which are all divisible by . However, when is odd, we will have a middle term that is not paired with any other terms, which is not necessarily divisible by . Thus, we have two cases:
is even
If is even, all terms that are greater than pair, as there are an even number of terms that are greater than . Therefore, all we need in order for the entire sequence to have a remainder of when divided by is to have a remainder of when divided by .
Evaluating as , all we need to be true is or that Thus, will be divisible by where . As is prime, must be equal to either or . If , we have that , which is not greater than or equal to , so that solution is extraneous.
If , we have that , which is , so one of our solutions is , and we are done with our first case.
is odd
If 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 . Therefore, as all other terms that are pair, the requirement that we have is
Calculating and simplifying, we have that
Now, we multiply both sides by . 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
As we know that is divisible by , what we need now is
We now check each solution to see whether it works.
If , would be less than , so none of these solutions work. If , 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 , , or . We plug these values into the congruence before we multiplied both sides by to check.
If , we would need Calculating and factoring out , we have that As the right parenthesis is odd and , we know that this solution works, so we have another solution: .
If , we would need As the left hand side is odd, but all multiples of is even, this solution is therefore extraneous.
If , we would need
Again, the left hand side is odd, and all multiples of are even, so this solution is extraneous.
Therefore, our final answer is .
~ CoolCarsOnTheRun
Solution 2 (w/ formula)
Let . Then we have So, . Testing, the cases, only fails. This leaves .
Solution 3 (Official MAA 1)
The sum of the cubes from 1 to is For this to be equal to for some integer , it must be thatsoBut Thus is congruent to both and which implies that divides . Because , the only choices for are and Checking all three cases verifies that and work, but does not. The requested sum is .
Solution 4 (Official MAA 2)
The sum of the cubes of the integers from through iswhich, when divided by , has quotientwith remainder If is not congruent to , then is an integer, andso divides , and . If , then is half of an integer, and letting for some integer givesThus divides . It follows that , and . The requested sum is .
Solution 5
Using the formula for , Since divided by has a remainder of , Using the rules of modular arithmetic, Expanding the left hand side, This means that is divisible by .
Dividing polynomials,
Note that and (because the remainder when dividing by is , so must be greater than ), so all options can be eliminated.
Checking all 3 cases, and work; fails.
Therefore, the answer is .
~ {TSun} ~
Solution 6 (similar ideas to Solution 1, but faster)
As before, we note that Thus, we can pair up the terms from to and cancel them. We have to deal with two cases:
If is even, then as there are an even number of these terms and they pair and cancel. We thus get or which yields
If is odd, then Letting yields However, this means that is divisible by so Plugging this back into yields in the latter case.
Thus, the sum of all possible is just
- ccx09
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 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
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