2006 AIME I Problems/Problem 5
Contents
Problem
The number 
can be written as 
where 
and 
are positive integers. Find 
.
Solution 1
We begin by equating the two expressions:
![\[a\sqrt{2}+b\sqrt{3}+c\sqrt{5} = \sqrt{104\sqrt{6}+468\sqrt{10}+144\sqrt{15}+2006}\]](http://latex.artofproblemsolving.com/5/6/7/56755e40101b654cd62aaaf4f16dc4760f70b9c1.png)
Squaring both sides yields:
![\[2ab\sqrt{6} + 2ac\sqrt{10} + 2bc\sqrt{15} + 2a^2 + 3b^2 + 5c^2 = 104\sqrt{6}+468\sqrt{10}+144\sqrt{15}+2006\]](http://latex.artofproblemsolving.com/3/b/a/3ba9d82307df69713fdd1fffeaf5e555a12725c5.png)
Since 
, 
, and are integers, we can match coefficients:
\[2ab\sqrt{6} &=& 104\sqrt{6} \\
2ac\sqrt{10} &=& 468\sqrt{10} \\
2bc\sqrt{15} &=& 144\sqrt{15}\\
2a^2 + 3b^2 + 5c^2 &=& 2006\] (Error compiling LaTeX. Unknown error_msg)
Solving the first three equations gives:
Multiplying these equations gives 
.
Solution 2
We realize that the quantity under the largest radical is a perfect square and attempt to rewrite the radicand as a square. Start by setting 
, 
, and 
. Since
![\[(px+qy+rz)^2=p^2x^2+q^2y^2+r^2z^2+2(pqxy+prxz+qryz)\]](http://latex.artofproblemsolving.com/2/e/f/2ef72c62a6e6b8b5597eddd75764f62195081bdc.png)
we attempt to rewrite the radicand in this form:
![\[2006+2(52xy+234xz+72yz)\]](http://latex.artofproblemsolving.com/a/7/3/a7302c22b2e8ea1e719616b856e4bb1f31a2723a.png)
Factoring, we see that 
, 
, and 
. Setting 
, 
, and 
, we see that
\[2006=13^2\cdotx^2+4^2\cdoty^2+18^2\cdotz^2=169\cdot2+16\cdot3+324\cdot5\] (Error compiling LaTeX. Unknown error_msg)
so our numbers check. Thus 
. Square rooting gives us 
and our answer is 
See also
| 2006 AIME I (Problems • Answer Key • Resources) | ||
| Preceded by Problem 4 |
Followed by Problem 6 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
| All AIME Problems and Solutions | ||
