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Difference between revisions of "Mock AIME 2 2006-2007 Problems/Problem 1"

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== Problem ==
 
== Problem ==
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A positive integer is called a dragon if it can be [[partition]]ed into four positive integers <math>\displaystyle a,b,c,</math> and <math>\displaystyle d</math> such that <math>\displaystyle a+4=b-4=4c=d/4.</math> Find the smallest dragon.
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A [[positive integer]] is called a ''dragon'' if it can be written as the sum of four positive integers <math>a,b,c,</math> and <math>d</math> such that <math>a+4=b-4=4c=d/4.</math> Find the smallest dragon.
  
 
==Solution==
 
==Solution==
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{{solution}}
 
  
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From <math>4c = \frac{d}4</math> we have that 16 [[divisor | divides]] <math>d</math>.  From <math>a + 4 = \frac d4</math> we have <math>d \geq 20</math>.  Minimizing <math>d</math> minimizes <math>a, b</math> and <math>c</math> and consequently minimizes our dragon.  The smallest possible choice is <math>d = 32</math>, from which <math>a = 4, b = 12</math> and <math>c = 2</math> so our desired number is <math>a + b + c + d = 4 + 12 + 2 + 32 = 050</math>.
  
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*[[Mock AIME 2 2006-2007/Problem 2 | Next Problem]]
 
  
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*[[Mock AIME 2 2006-2007]]
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==Solution2==
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Note: a+b+c+d=a+a+8+(a+4)/4+4a+16
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=(25a+100)/4
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= 25(a+4)/4 . Therefore, we know that a+4/4 has to be an integer. since a>0, the smaller integer we can have is 2, so a=4.
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Therefore, the smallest value is 50.
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~Please convert to latex
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==See Also==
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{{Mock AIME box|year=2006-2007|n=2|before=First Question|num-a=2}}
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[[Category:Introductory Number Theory Problems]]
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[[Category:Introductory Algebra Problems]]

Latest revision as of 00:27, 11 April 2024

Problem

A positive integer is called a dragon if it can be written as the sum of four positive integers $a,b,c,$ and $d$ such that $a+4=b-4=4c=d/4.$ Find the smallest dragon.

Solution

From $4c = \frac{d}4$ we have that 16 divides $d$. From $a + 4 = \frac d4$ we have $d \geq 20$. Minimizing $d$ minimizes $a, b$ and $c$ and consequently minimizes our dragon. The smallest possible choice is $d = 32$, from which $a = 4, b = 12$ and $c = 2$ so our desired number is $a + b + c + d = 4 + 12 + 2 + 32 = 050$.


Solution2

Note: a+b+c+d=a+a+8+(a+4)/4+4a+16 =(25a+100)/4 = 25(a+4)/4 . Therefore, we know that a+4/4 has to be an integer. since a>0, the smaller integer we can have is 2, so a=4. Therefore, the smallest value is 50.

~Please convert to latex

See Also

Mock AIME 2 2006-2007 (Problems, Source)
Preceded by
First Question 		Followed by
Problem 2
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
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