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

(Created page with '== Problem 3 == 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 <ma…')
 
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== Solution ==
 
== Solution ==
  
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I will type it right now !!!
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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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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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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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<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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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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<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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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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<br/>
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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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Sum <math>=4</math>
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<br/>
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summing all this we have <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}}

Revision as of 17:42, 3 April 2010

Problem 3

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 make $b$ anyway from $n+1$ to $20$ and make $a$ $b-n$.

Thus, the product is $(1^19)(2^18)\cdots(19^1)$ (some people may recognize it as $19!18!\cdots1!$.)

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 all this we have $\boxed{150}$

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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