Difference between revisions of "2006 AIME I Problems/Problem 5"
(the second half really isn't necessary, and taking sqrts aren't hard anyway..) |
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The number <math> \sqrt{104\sqrt{6}+468\sqrt{10}+144\sqrt{15}+2006}</math> can be written as <math> a\sqrt{2}+b\sqrt{3}+c\sqrt{5}, </math> where <math> a, b, </math> and <math> c </math> are [[positive]] [[integer]]s. Find <math>abc</math>. | The number <math> \sqrt{104\sqrt{6}+468\sqrt{10}+144\sqrt{15}+2006}</math> can be written as <math> a\sqrt{2}+b\sqrt{3}+c\sqrt{5}, </math> where <math> a, b, </math> and <math> c </math> are [[positive]] [[integer]]s. Find <math>abc</math>. | ||
− | == Solution == | + | == Solution 1 == |
We begin by [[equate | equating]] the two expressions: | We begin by [[equate | equating]] the two expressions: | ||
Line 72: | Line 72: | ||
<math>abc=\boxed{936}</math> | <math>abc=\boxed{936}</math> | ||
--> | --> | ||
+ | |||
+ | == 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 <math>x=\sqrt{2}</math>, <math>y=\sqrt{3}</math>, and <math>z=\sqrt{5}</math>. Since | ||
+ | |||
+ | <cmath>(px+qy+rz)^2=p^2x^2+q^2y^2+r^2z^2+2(pqxy+prxz+qryz)</cmath> | ||
+ | |||
+ | we attempt to rewrite the radicand in this form: | ||
+ | |||
+ | <cmath>2006+2(52xy+234xz+72yz)</cmath> | ||
+ | |||
+ | Factoring, we see that <math>52=13\cdot4</math>, <math>234=13\cdot18</math>, and <math>72=4\cdot18</math>. Setting <math>p=13</math>, <math>q=4</math>, and <math>r=18</math>, we see that | ||
+ | |||
+ | <cmath>2006=13^2\cdotx^2+4^2\cdoty^2+18^2\cdotz^2=169\cdot2+16\cdot3+324\cdot5</cmath> | ||
+ | |||
+ | so our numbers check. Thus <math>104\sqrt{2}+468\sqrt{3}+144\sqrt{5}=(13\sqrt{2}+4\sqrt{3}+18\sqrt{5})^2</math>. Square rooting gives us <math>13\sqrt{2}+4\sqrt{3}+18\sqrt{5}</math> and our answer is <math>13\cdot4\cdot18=\boxed{936}</math> | ||
+ | |||
== See also == | == See also == | ||
{{AIME box|year=2006|n=I|num-b=4|num-a=6}} | {{AIME box|year=2006|n=I|num-b=4|num-a=6}} | ||
[[Category:Intermediate Algebra Problems]] | [[Category:Intermediate Algebra Problems]] |
Revision as of 01:35, 13 July 2012
Contents
Problem
The number can be written as
where
and
are positive integers. Find
.
Solution 1
We begin by equating the two expressions:
Squaring both sides yields:
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
we attempt to rewrite the radicand in this form:
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 |