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Difference between revisions of "2010 AIME II Problems/Problem 3"

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== Problem 3 ==
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== Problem ==
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Let <math>K</math> be the product of all factors <math>(b-a)</math> (not necessarily distinct) where <math>a</math> and <math>b</math> are integers satisfying <math>1\le a < b \le 20</math>. Find the greatest positive integer <math>n</math> such that <math>2^n</math> divides <math>K</math>.
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Let <math>K</math> be the product of all factors <math>(b-a)</math> (not necessarily distinct) where <math>a</math> and <math>b</math> are integers satisfying <math>1\le a < b \le 20</math>. Find the greatest positive [[integer]] <math>n</math> such that <math>2^n</math> divides <math>K</math>.
  
 
== Solution ==
 
== Solution ==
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In general, there are <math>20-n</math> pairs of integers <math>(a, b)</math> that differ by <math>n</math> because we can let <math>b</math> be any integer from <math>n+1</math> to <math>20</math> and set <math>a</math> equal to <math>b-n</math>. Thus, the product is <math>(1^{19})(2^{18})\cdots(19^1)</math> (or alternatively, <math>19! \cdot 18! \cdots 1!</math>.)
  
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In general, there are <math>20-n</math> pairs of integers <math>(a, b)</math> that differ by <math>n</math> because we can make <math>b</math> anyway from <math>n+1</math> to <math>20</math> and make <math>a</math> <math>b-n</math>.
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When we count the number of factors of <math>2</math>, we have 4 groups, factors that are divisible by <math>2</math> at least once, twice, three times and four times.
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Thus, the product is <math>(1^{19})(2^{18})\cdots(19^1)</math> (some people may recognize it as <math>19!18!\cdots1!</math>.)
 
  
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When we count the number of factors of <math>2</math>, we have 4 groups, factors that are divisible by <math>2</math> at least once, twice, three times and four times.
 
 
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<br/>
  
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Number that are divisible by <math>2</math> at least once: <math>2, 4, \cdots, 18</math>
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*Number that are divisible by <math>2</math> at least once: <math>2, 4, \cdots, 18</math>
  
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Exponent corresponding to each one of them <math>18, 16, \cdots 2</math>
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:Exponent corresponding to each one of them <math>18, 16, \cdots 2</math>
  
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Sum <math>=2+4+\cdots+18=\frac{(20)(9)}{2}=90</math>
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:Sum <math>=2+4+\cdots+18=\frac{(20)(9)}{2}=90</math>
 
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<br/>
  
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Number that are divisible by <math>2</math> at least twice: <math>4, 8, \cdots, 16</math>
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*Number that are divisible by <math>2</math> at least twice: <math>4, 8, \cdots, 16</math>
  
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Exponent corresponding to each one of them <math>16, 12, \cdots 4</math>
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:Exponent corresponding to each one of them <math>16, 12, \cdots 4</math>
  
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Sum <math>=4+8+\cdots+16=\frac{(20)(4)}{2}=40</math>  
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:Sum <math>=4+8+\cdots+16=\frac{(20)(4)}{2}=40</math>  
  
 
<br/>
 
<br/>
  
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Number that are divisible by <math>2</math> at least three times: <math>8,16</math>
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*Number that are divisible by <math>2</math> at least three times: <math>8,16</math>
  
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Exponent corresponding to each one of them <math>12, 4</math>
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:Exponent corresponding to each one of them <math>12, 4</math>
  
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Sum <math>=12+4=16</math>  
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:Sum <math>=12+4=16</math>  
  
 
<br/>
 
<br/>
  
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Number that are divisible by <math>2</math> at least four times: <math>16</math>
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*Number that are divisible by <math>2</math> at least four times: <math>16</math>
  
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Exponent corresponding to each one of them <math>4</math>
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:Exponent corresponding to each one of them <math>4</math>
  
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Sum <math>=4</math>  
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:Sum <math>=4</math>  
  
 
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summing all this we have <math>\boxed{150}</math>
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Summing these give an answer of <math>\boxed{150}</math>.
  
 
== See also ==
 
== See also ==
 
{{AIME box|year=2010|num-b=2|num-a=4|n=II}}
 
{{AIME box|year=2010|num-b=2|num-a=4|n=II}}
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[[Category:Intermediate Number Theory Problems]]

Revision as of 10:27, 6 April 2010

Problem

Let $K$ be the product of all factors $(b-a)$ (not necessarily distinct) where $a$ and $b$ are integers satisfying $1\le a < b \le 20$. Find the greatest positive integer $n$ such that $2^n$ divides $K$.

Solution

In general, there are $20-n$ pairs of integers $(a, b)$ that differ by $n$ because we can let $b$ be any integer from $n+1$ to $20$ and set $a$ equal to $b-n$. Thus, the product is $(1^{19})(2^{18})\cdots(19^1)$ (or alternatively, $19! \cdot 18! \cdots 1!$.)

When we count the number of factors of $2$, we have 4 groups, factors that are divisible by $2$ at least once, twice, three times and four times.


  • Number that are divisible by $2$ at least once: $2, 4, \cdots, 18$
Exponent corresponding to each one of them $18, 16, \cdots 2$
Sum $=2+4+\cdots+18=\frac{(20)(9)}{2}=90$


  • Number that are divisible by $2$ at least twice: $4, 8, \cdots, 16$
Exponent corresponding to each one of them $16, 12, \cdots 4$
Sum $=4+8+\cdots+16=\frac{(20)(4)}{2}=40$


  • Number that are divisible by $2$ at least three times: $8,16$
Exponent corresponding to each one of them $12, 4$
Sum $=12+4=16$


  • Number that are divisible by $2$ at least four times: $16$
Exponent corresponding to each one of them $4$
Sum $=4$


Summing these give an answer of $\boxed{150}$.

See also

2010 AIME II (Problems • Answer Key • Resources)
Preceded by
Problem 2 		Followed by
Problem 4
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions
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