1995 AIME Problems/Problem 2
Contents
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:
Hrm... we know that and
are reciprocals, so let
. Then we have
. Multiplying by
and simplifying gives us
, as shown above.
By Vieta's formulas, the sum of the possible values of is
. This means that the roots
and
that satisfy the original equation also satisfy
We can combine these logs to get
, or
. Finally, we find this value mod
, which is easy.
, so our answer is
.
See also
1995 AIME (Problems • Answer Key • Resources) | ||
Preceded by Problem 1 |
Followed by Problem 3 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
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