1995 AIME Problems/Problem 2
Problem
Find the last three digits of the product of the positive roots of .
Solution 1
Taking the (logarithm) of both sides and then moving to one side yields the quadratic equation . Applying the quadratic formula yields that . Thus, the product of the two roots (both of which are positive) is , making the solution .
Solution 2
Instead of taking , we take of both sides and simplify:
We know that and are reciprocals, so let . Then we have . Multiplying by and simplifying gives us , as shown above.
Because , . By the quadratic formula, the two roots of our equation are . This means our two roots in terms of are and Multiplying these gives
, so our answer is .
Solution 3
Let . Rewriting the equation in terms of , we have Thus, the product of the positive roots is , so the last three digits are .
See also
1995 AIME (Problems • Answer Key • Resources) | ||
Preceded by Problem 1 |
Followed by Problem 3 | |
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