1989 AIME Problems/Problem 1
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 | |
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All AIME Problems and Solutions |