Difference between revisions of "2006 AIME I Problems/Problem 5"
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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: | ||
− | < | + | <cmath> a\sqrt{2}+b\sqrt{3}+c\sqrt{5} = \sqrt{104\sqrt{6}+468\sqrt{10}+144\sqrt{15}+2006}</cmath> |
− | Squaring both sides | + | Squaring both sides yields: |
− | < | + | <cmath> 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 </cmath> |
− | Since <math>a</math>, <math>b</math>, and <math>c</math> are integers: | + | Since <math>a</math>, <math>b</math>, and <math>c</math> are integers, we can match coefficients: |
− | + | <cmath> | |
+ | \begin{align*} | ||
+ | 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 | ||
+ | \end{align*} | ||
+ | </cmath> | ||
− | + | Solving the first three equations gives: | |
+ | <cmath>\begin{eqnarray*}ab &=& 52\\ | ||
+ | ac &=& 234\\ | ||
+ | bc &=& 72 \end{eqnarray*}</cmath> | ||
− | + | Multiplying these equations gives <math> (abc)^2 = 52 \cdot 234 \cdot 72 = 2^63^413^2 \Longrightarrow abc = \boxed{936}</math>. | |
− | + | <!-- | |
− | + | Since this is the AIME and you do not have a calculator solving <math> abc = \sqrt{52 \cdot 234 \cdot 72} = 936</math> might prove difficult. | |
− | + | So instead use the three equations given above. | |
<math> ab = 52 </math> | <math> ab = 52 </math> | ||
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<math> bc = 72 </math> | <math> bc = 72 </math> | ||
− | + | Thus <math> a = 52/b = 234/c </math> | |
+ | |||
+ | <math> 52c = 234b </math> | ||
+ | |||
+ | <math> c = 234/52b </math> | ||
+ | |||
+ | <math> c = 9/2b </math> | ||
+ | |||
+ | Plugging into last equation leads to: | ||
+ | |||
+ | <math> 9/2b^2 = 72 </math> | ||
+ | |||
+ | <math> b = 4 </math> | ||
+ | |||
+ | Plugging into others you get | ||
+ | |||
+ | <math>a=13</math> | ||
+ | |||
+ | <math>b=4</math> | ||
− | <math> | + | <math>c=18</math> |
− | + | Much easier than taking crazy square roots without a calculator! | |
If it was required to solve for each variable, dividing the product of the three variables by the product of any two variables would yield the third variable. Doing so yields: | If it was required to solve for each variable, dividing the product of the three variables by the product of any two variables would yield the third variable. Doing so yields: | ||
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<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=13^2x^2+4^2y^2+18^2z^2=169\cdot2+16\cdot3+324\cdot5</cmath> | ||
+ | |||
+ | so our numbers check. Thus <math>104\sqrt{2}+468\sqrt{3}+144\sqrt{5}+2006=(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 == | ||
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[[Category:Intermediate Algebra Problems]] | [[Category:Intermediate Algebra Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 18:24, 10 March 2015
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:
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
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.