Difference between revisions of "1989 AIME Problems/Problem 1"
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===Solution 4=== | ===Solution 4=== | ||
− | Similar to Solution 1 above, call the consecutive integers <math>(n-\frac{3}{2}), (n-\frac{1}{2}), (n+\frac{1}{2}), (n+\frac{3}{2})</math> to make use of symmetry. Note that <math>n</math> itself is not an integer - in this case, <math>n = 29.5</math>. The expression becomes <math>\sqrt{(n-\frac{3}{2})(n + \frac{3}{2})(n - \frac{1}{2})(n + \frac{1}{2}) + 1}</math>. Distributing each pair of difference of squares first, and then distributing the two resulting quadratics and adding the constant, gives <math>\sqrt{n^4 - \frac{5}{2}n^2 + \frac{25}{16}}</math>. The inside is a perfect square trinomial, since <math>b^2 = 4ac</math>. It's equal to <math>\sqrt{(n^2 - \frac{5}{4})^2}</math>, which simplifies to <math>n^2 - \frac{5}{4}</math>. You can plug in the value of <math>n</math> from there, or further simplify to <math>(n - \frac{1}{2})(n + \frac{1}{2} - 1 | + | Similar to Solution 1 above, call the consecutive integers <math>(n-\frac{3}{2}), (n-\frac{1}{2}), (n+\frac{1}{2}), (n+\frac{3}{2})</math> to make use of symmetry. Note that <math>n</math> itself is not an integer - in this case, <math>n = 29.5</math>. The expression becomes <math>\sqrt{(n-\frac{3}{2})(n + \frac{3}{2})(n - \frac{1}{2})(n + \frac{1}{2}) + 1}</math>. Distributing each pair of difference of squares first, and then distributing the two resulting quadratics and adding the constant, gives <math>\sqrt{n^4 - \frac{5}{2}n^2 + \frac{25}{16}}</math>. The inside is a perfect square trinomial, since <math>b^2 = 4ac</math>. It's equal to <math>\sqrt{(n^2 - \frac{5}{4})^2}</math>, which simplifies to <math>n^2 - \frac{5}{4}</math>. You can plug in the value of <math>n</math> from there, or further simplify to <math>(n - \frac{1}{2})(n + \frac{1}{2}) - 1</math>, which is easier to compute. Either way, plugging in <math>n=29.5</math> gives <math>\boxed{869}</math>. |
== See also == | == See also == |
Revision as of 08:00, 21 August 2012
Contents
Problem
Compute .
Solution
Solution 1
Let's call our four consecutive integers . Notice that . Thus, .
Solution 2
Note that the four numbers to multiply are symmetric with the center at . Multiply the symmetric pairs to get and . Now clearly .
Solution 3
The last digit under the radical is , so the square root must either end in or , since means . Additionally, the number must be near , narrowing the reasonable choices to and .
Continuing the logic, the next-to-last digit under the radical is the same as the last digit of , which is . Quick computation shows that ends in , while ends in . Thus, the answer is .
Solution 4
Similar to Solution 1 above, call the consecutive integers to make use of symmetry. Note that itself is not an integer - in this case, . The expression becomes . Distributing each pair of difference of squares first, and then distributing the two resulting quadratics and adding the constant, gives . The inside is a perfect square trinomial, since . It's equal to , which simplifies to . You can plug in the value of from there, or further simplify to , which is easier to compute. Either way, plugging in gives .
See also
1989 AIME (Problems • Answer Key • Resources) | ||
Preceded by First Question |
Followed by Problem 2 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |