Difference between revisions of "1995 AIME Problems/Problem 2"
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<math>\sqrt{1995}x^{\log_{1995}x}=x^2</math>. | <math>\sqrt{1995}x^{\log_{1995}x}=x^2</math>. | ||
− | + | ==Solution 1== | |
Taking the <math>\log_{1995}</math> ([[logarithm]]) of both sides and then moving to one side yields the [[quadratic equation]] <math>2(\log_{1995}x)^2 - 4(\log_{1995}x) + 1 = 0</math>. Applying the [[quadratic formula]] yields that <math>\log_{1995}x = 1 \pm \frac{\sqrt{2}}{2}</math>. Thus, the product of the two roots (both of which are positive) is <math>1995^{1+\sqrt{2}/2} \cdot 1995^{1 - \sqrt{2}/2} = 1995^2</math>, making the solution <math>(2000-5)^2 \equiv \boxed{025} \pmod{1000}</math>. | Taking the <math>\log_{1995}</math> ([[logarithm]]) of both sides and then moving to one side yields the [[quadratic equation]] <math>2(\log_{1995}x)^2 - 4(\log_{1995}x) + 1 = 0</math>. Applying the [[quadratic formula]] yields that <math>\log_{1995}x = 1 \pm \frac{\sqrt{2}}{2}</math>. Thus, the product of the two roots (both of which are positive) is <math>1995^{1+\sqrt{2}/2} \cdot 1995^{1 - \sqrt{2}/2} = 1995^2</math>, making the solution <math>(2000-5)^2 \equiv \boxed{025} \pmod{1000}</math>. | ||
− | + | ==Solution 2== | |
Instead of taking <math>\log_{1995}</math>, we take <math>\log_x</math> of both sides and simplify: | Instead of taking <math>\log_{1995}</math>, we take <math>\log_x</math> of both sides and simplify: | ||
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<math>1995^2\pmod{1000}\equiv 995^2\pmod{1000}\equiv (-5)^2\pmod{1000}\equiv 25\pmod{1000}</math>, so our answer is <math>\boxed{025}</math>. | <math>1995^2\pmod{1000}\equiv 995^2\pmod{1000}\equiv (-5)^2\pmod{1000}\equiv 25\pmod{1000}</math>, so our answer is <math>\boxed{025}</math>. | ||
− | + | ==Solution 3== | |
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> |
Latest revision as of 00:40, 29 January 2021
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.