Difference between revisions of "1992 AIME Problems/Problem 4"
(→Solution) |
|||
| Line 15: | Line 15: | ||
After canceling out the <math>n!</math> in the numerator and the <math>(n-r)!r!</math> in the denominator, we get <math>\frac{20r}{n-r+1} = 15 = \frac{12(n-r)}{r+1}</math>. Setting the first equation to <math>15</math> and the third equation to <math>15</math>, we get a system that is solvable. We have: <cmath>20r = 15n - 15r + 15 \Rightarrow 35r - 15n = 15</cmath> <cmath>15r + 15 = 12n - 12r \Rightarrow 27r - 12n = -15</cmath> | After canceling out the <math>n!</math> in the numerator and the <math>(n-r)!r!</math> in the denominator, we get <math>\frac{20r}{n-r+1} = 15 = \frac{12(n-r)}{r+1}</math>. Setting the first equation to <math>15</math> and the third equation to <math>15</math>, we get a system that is solvable. We have: <cmath>20r = 15n - 15r + 15 \Rightarrow 35r - 15n = 15</cmath> <cmath>15r + 15 = 12n - 12r \Rightarrow 27r - 12n = -15</cmath> | ||
| − | Solving these equations, we get that <math>r = 27</math> and <math>n = 62</math>. Our goal is to find which row of Pascal's triangle this ratio occurs, or in other words find what n is, which we conclude to be <math>\boxed{62}</math> | + | Solving these equations, we get that <math>r = 27</math> and <math>n = 62</math>. Our goal is to find which row of Pascal's triangle this ratio occurs, or in other words find what n is, which we conclude to be <math>\boxed{62}.</math> |
Revision as of 12:31, 26 March 2020
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 
?
Solution
In Pascal's Triangle, we know that the binomial coefficients of the 
th row are 
. Let our row be the th row such that the three consecutive entries are 
, 
and 
. (We consider 
because it cancels stuff out easier)
Consider what equals to in a more explicit form. It equals 
.
Now consider what it means to have three consecutive entries occurring in the ratio 
. It means that we will have 
. Note that the order of the ratio does not matter, as ascending from one side of Pascal's triangle is equivalent to descending from the opposite side of Pascal's triangle. We can multiply by a LCM of 
to further simplify the problem into 
Using the more explicit form of , we see that this equivalence function collapses into 
(all of which is given by plugging in into )
After canceling out the 
in the numerator and the 
in the denominator, we get 
. Setting the first equation to 
and the third equation to , we get a system that is solvable. We have: ![\[20r = 15n - 15r + 15 \Rightarrow 35r - 15n = 15\]](http://latex.artofproblemsolving.com/c/9/1/c917a55a2e3f8a98e61522f3364c7370e28f0f5c.png)
![\[15r + 15 = 12n - 12r \Rightarrow 27r - 12n = -15\]](http://latex.artofproblemsolving.com/4/7/7/477f3484e570977d70babe6f5a68951b95971e65.png)
Solving these equations, we get that 
and 
. Our goal is to find which row of Pascal's triangle this ratio occurs, or in other words find what n is, which we conclude to be 
| 1992 AIME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 3 |
Followed by Problem 5 | |
| 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. 
