Difference between revisions of "1995 AIME Problems/Problem 2"
Flyhawkeye (talk | contribs) (Added Solution 3) |
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Let <math>y=\log_{1995}x</math>. Rewriting the equation in terms of <math>y</math>, we have | Let <math>y=\log_{1995}x</math>. Rewriting the equation in terms of <math>y</math>, we have | ||
<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 | + | <cmath> 1995^{y^2+\frac{1}{2}}=1995^{2y}</cmath> |
<cmath> y^2-\frac{1}{2}=2y</cmath> | <cmath> y^2-\frac{1}{2}=2y</cmath> | ||
− | <cmath> 2y^2-4y | + | <cmath> 2y^2-4y+1=0</cmath> |
− | <cmath> y=\frac{4\pm\sqrt{16-\left(4\right)\left(2\right)\left( | + | <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> |
− | Thus, the product of the positive roots is <math>\left(1995^{\frac{2+\sqrt{ | + | Thus, the product of the positive roots is <math>\left(1995^{\frac{2+\sqrt{8}}{2}}\right)\left(1995^{\frac{2-\sqrt{8}}{2}}\right)=1995^2=\left(2000-5\right)^2</math>, so the last three digits are <math>\boxed{025}</math>. |
== See also == | == See also == |
Revision as of 22:51, 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.