Difference between revisions of "1995 AIME Problems/Problem 2"
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<cmath> \sqrt{1995}\left(1995^y\right)^y=1995^{2y}</cmath> | <cmath> \sqrt{1995}\left(1995^y\right)^y=1995^{2y}</cmath> | ||
<cmath> 1995^{y^2+\frac{1}{2}}=1995^{2y}</cmath> | <cmath> 1995^{y^2+\frac{1}{2}}=1995^{2y}</cmath> | ||
− | <cmath> y^2 | + | <cmath> y^2+\frac{1}{2}=2y</cmath> |
<cmath> 2y^2-4y+1=0</cmath> | <cmath> 2y^2-4y+1=0</cmath> | ||
<cmath> y=\frac{4\pm\sqrt{16-\left(4\right)\left(2\right)\left(1\right)}}{4}=\frac{4\pm\sqrt{8}}{4}=\frac{2\pm\sqrt{8}}{2}</cmath> | <cmath> y=\frac{4\pm\sqrt{16-\left(4\right)\left(2\right)\left(1\right)}}{4}=\frac{4\pm\sqrt{8}}{4}=\frac{2\pm\sqrt{8}}{2}</cmath> |
Revision as of 23:52, 12 August 2018
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 | |
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.