Difference between revisions of "2017 AIME II Problems/Problem 6"
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The solution to the initial quadratic in <math>n</math> must then be <math>\frac{-85 \pm \sqrt{85^2 - 4(2017 - x^2)}}{2}</math>. Noticing that for each value of <math>x</math> that has rational solutions for <math>n</math>, the corresponding value of the square root of the discriminant is <math>y</math>, the formula can be rewritten as <math>n = \frac{-85 \pm y}{2}</math>. One solution is <math>\frac{421 - 85}{2} = 168</math> and the other solution is <math>\frac{139 - 85}{2} = 27</math>. Thus the answer is <math>168 + 27 = \boxed{195}</math> as both rational solutions are integers. | The solution to the initial quadratic in <math>n</math> must then be <math>\frac{-85 \pm \sqrt{85^2 - 4(2017 - x^2)}}{2}</math>. Noticing that for each value of <math>x</math> that has rational solutions for <math>n</math>, the corresponding value of the square root of the discriminant is <math>y</math>, the formula can be rewritten as <math>n = \frac{-85 \pm y}{2}</math>. One solution is <math>\frac{421 - 85}{2} = 168</math> and the other solution is <math>\frac{139 - 85}{2} = 27</math>. Thus the answer is <math>168 + 27 = \boxed{195}</math> as both rational solutions are integers. | ||
− | ==Solution 4== | + | ==Solution 4 (Bounding)== |
Notice that <math>(n+42)^2= n^2+84n+1764</math>. Also note that <math>(n+45)^2= n^2+90n+2025</math>. Thus, <cmath>(n+42)^2< n^2+85n+2017<(n+45)^2</cmath> where <math>n^2+85n+2017</math> is a perfect square. Hence,<cmath>n^2+85n+2017= (n+43)^2</cmath> or <cmath>n^2+85n+2017= (n+44)^2.</cmath> Solving the two equations yields the two solutions <math>n= 168, 27</math>. Therefore, our answer is <math>\boxed{195}</math>. | Notice that <math>(n+42)^2= n^2+84n+1764</math>. Also note that <math>(n+45)^2= n^2+90n+2025</math>. Thus, <cmath>(n+42)^2< n^2+85n+2017<(n+45)^2</cmath> where <math>n^2+85n+2017</math> is a perfect square. Hence,<cmath>n^2+85n+2017= (n+43)^2</cmath> or <cmath>n^2+85n+2017= (n+44)^2.</cmath> Solving the two equations yields the two solutions <math>n= 168, 27</math>. Therefore, our answer is <math>\boxed{195}</math>. |
Revision as of 20:41, 27 September 2019
Contents
Problem
Find the sum of all positive integers such that is an integer.
Solution 1
Manipulating the given expression, . The expression under the radical must be an square number for the entire expression to be an integer, so . Rearranging, . By difference of squares, . It is easy to check that those are all the factor pairs of 843. Considering each factor pair separately, is found to be and . The two values of that satisfy one of the equations are and . Summing these together, the answer is .
Solution 2
Clearly, the result when is plugged into the given expression is larger than itself. Let be the positive difference between that result and , so that . Squaring both sides and canceling the terms gives . Combining like terms, , so
Since is positive, there are two cases, which are simple (luckily). Remembering that is a positive integer, then and are either both positive or both negative. The smallest value for which is 45, which makes the denominator, and the entire expression, negative. Evaluating the other case where numerator and denominator are both negative, then we have that (from the numerator) and , which means . This only gives two solutions, . Plugging these into the expression for , we find that they result in 27 and 168, which both satisfy the initial question. Therefore, the answer is .
Solution 3 (Abuse the discriminant)
Let the integer given by the square root be represented by . Then . For this to have rational solutions for (checking whether they are integers is done later), the discriminant of this quadratic must be a perfect square. (This can be easily shown using the quadratic formula.)
Thus, for some integer . Then . Rearranging this equation yields that . Noticing that there are 2 factor pairs of , namely, and , there are 2 systems to solve for and that create rational . These yield solutions of and .
The solution to the initial quadratic in must then be . Noticing that for each value of that has rational solutions for , the corresponding value of the square root of the discriminant is , the formula can be rewritten as . One solution is and the other solution is . Thus the answer is as both rational solutions are integers.
Solution 4 (Bounding)
Notice that . Also note that . Thus, where is a perfect square. Hence, or Solving the two equations yields the two solutions . Therefore, our answer is .
Solution 5 (Using factors)
Let the expression be equal to . This expression can be factored into . Then square both sides, and the expression becomes . We have a difference of two squares. Rearranging, we have . By inspection, the only possible values for are 0 and 1. When , we must have . Therefore, is a solution. When we have , so . Plugging this back to (since ), we find that . Thus, the answer is .
-RootThreeOverTwo
Solution 6
Ignore the square root for now. This expression can be factored into . Just by inspection, when , the expression becomes , so is a solution. Proceed as Solution 5 to find the other solution(s).
See Also
2017 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 5 |
Followed by Problem 7 | |
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.