Difference between revisions of "1992 AIME Problems/Problem 4"

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In Pascal's Triangle, we know that the binomial coefficients of the <math>n</math>th row are <math>n\choose 0</math>, <math>n\choose1</math>, ..., <math>n\choose n</math>. Let our row be the <math>n</math>th row such that the three consecutive entries are <math>n \choose r</math>, <math>n \choose r+1</math> and <math>n \choose r+2</math>.  
 
In Pascal's Triangle, we know that the binomial coefficients of the <math>n</math>th row are <math>n\choose 0</math>, <math>n\choose1</math>, ..., <math>n\choose n</math>. Let our row be the <math>n</math>th row such that the three consecutive entries are <math>n \choose r</math>, <math>n \choose r+1</math> and <math>n \choose r+2</math>.  
  
After expanding and dividing one entry by another (to clean up the factorials), we see that <math>\displaystyle \frac 34=\frac{r+1}{n-r}</math> and <math>\displaystyle \frac45=\frac{r+2}{n-r-1}</math>. Solving, <math>\displaystyle n = 62</math>.
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After expanding and dividing one entry by another (to clean up the factorials), we see that <math>\frac 34=\frac{r+1}{n-r}</math> and <math>\frac45=\frac{r+2}{n-r-1}</math>. Solving, <math>n = 62</math>.
  
== See also ==
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{{AIME box|year=1992|num-b=1|num-a=3}}
* [[1992 AIME Problems/Problem 3 | Previous Problem]]
 
 
 
* [[1992 AIME Problems/Problem 5 | Next Problem]]
 
 
 
* [[1992 AIME Problems]]
 
  
 
[[Category:Intermediate Combinatorics Problems]]
 
[[Category:Intermediate Combinatorics Problems]]
 
[[Category:Intermediate Number Theory Problems]]
 
[[Category:Intermediate Number Theory Problems]]

Revision as of 21:57, 11 November 2007

Problem

In Pascal's Triangle, each entry is the sum of the two entries above it. In which row of Pascal's Triangle do three consecutive entries occur that are in the ratio $3: 4: 5$?

Solution

In Pascal's Triangle, we know that the binomial coefficients of the $n$th row are $n\choose 0$, $n\choose1$, ..., $n\choose n$. Let our row be the $n$th row such that the three consecutive entries are $n \choose r$, $n \choose r+1$ and $n \choose r+2$.

After expanding and dividing one entry by another (to clean up the factorials), we see that $\frac 34=\frac{r+1}{n-r}$ and $\frac45=\frac{r+2}{n-r-1}$. Solving, $n = 62$.

1992 AIME (ProblemsAnswer KeyResources)
Preceded by
Problem 1
Followed by
Problem 3
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