Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "1992 AIME Problems/Problem 4"
m (→‎Solution 1)
Line 42: Line 42:
 
<cmath> 28n - 35 = 27n+27 </cmath>
 
<cmath> 28n - 35 = 27n+27 </cmath>
 
<cmath> n = 62 </cmath>
 
<cmath> n = 62 </cmath>
−
We can also evaluate for <math>k</math>, and find that <math>k = \frac{3(62+1)}{7} = 27.</math> Since we want <math>n</math>, however, our final answer is <math>\boxed{062.}</math> ~LaTeX by ciceronii
+
We can also evaluate for <math>k</math>, and find that <math>k = \frac{3(62+1)}{7} = 27.</math> Since we want <math>n</math>, however, our final answer is <math>\boxed{062.}</math> ~<math>\LaTeX</math> by ciceronii
−
 
 
  
 
==Solution 2==
 
==Solution 2==

Revision as of 22:10, 17 July 2023

Problem

In Pascal's Triangle, each entry is the sum of the two entries above it. The first few rows of the triangle are shown below.

\[\begin{array}{c@{\hspace{8em}} c@{\hspace{6pt}}c@{\hspace{6pt}}c@{\hspace{6pt}}c@{\hspace{4pt}}c@{\hspace{2pt}} c@{\hspace{2pt}}c@{\hspace{2pt}}c@{\hspace{2pt}}c@{\hspace{3pt}}c@{\hspace{6pt}} c@{\hspace{6pt}}c@{\hspace{6pt}}c} \vspace{4pt} \text{Row 0: } & & & & & & & 1 & & & & & & \\\vspace{4pt} \text{Row 1: } & & & & & & 1 & & 1 & & & & & \\\vspace{4pt} \text{Row 2: } & & & & & 1 & & 2 & & 1 & & & & \\\vspace{4pt} \text{Row 3: } & & & & 1 & & 3 & & 3 & & 1 & & & \\\vspace{4pt} \text{Row 4: } & & & 1 & & 4 & & 6 & & 4 & & 1 & & \\\vspace{4pt} \text{Row 5: } & & 1 & & 5 & &10& &10 & & 5 & & 1 & \\\vspace{4pt} \text{Row 6: } & 1 & & 6 & &15& &20& &15 & & 6 & & 1 \end{array}\] In which row of Pascal's Triangle do three consecutive entries occur that are in the ratio $3 :4 :5$?

Solution 1

Consider what the ratio means. Since we know that they are consecutive terms, we can say \[\frac{\dbinom{n}{k-1}}{3} = \frac{\dbinom{n}{k}}{4} = \frac{\dbinom{n}{k+1}}{5}.\]

Taking the first part, and using our expression for $n$ choose $k$, \[\frac{n!}{3(k-1)!(n-k+1)!} = \frac{n!}{4k!(n-k)!}\] \[\frac{1}{3(k-1)!(n-k+1)!} = \frac{1}{4k!(n-k)!}\] \[\frac{1}{3(n-k+1)} = \frac{1}{4k}\] \[n-k+1 = \frac{4k}{3}\] \[n = \frac{7k}{3} - 1\] \[\frac{3(n+1)}{7} = k\] Then, we can use the second part of the equation. \[\frac{n!}{4k!(n-k)!} = \frac{n!}{5(k+1)!(n-k-1)!}\] \[\frac{1}{4k!(n-k)!} = \frac{1}{5(k+1)!(n-k-1)!}\] \[\frac{1}{4(n-k)} = \frac{1}{5(k+1)}\] \[\frac{4(n-k)}{5} = k+1\] \[\frac{4n}{5}-\frac{4k}{5} = k+1\] \[\frac{4n}{5} = \frac{9k}{5} +1.\] Since we know $k = \frac{3(n+1)}{7}$ we can plug this in, giving us \[\frac{4n}{5} = \frac{9\left(\frac{3(n+1)}{7}\right)}{5} +1\] \[4n = 9\left(\frac{3(n+1)}{7}\right)+5\] \[7(4n - 5) = 27n+27\] \[28n - 35 = 27n+27\] \[n = 62\] We can also evaluate for $k$, and find that $k = \frac{3(62+1)}{7} = 27.$ Since we want $n$, however, our final answer is $\boxed{062.}$ ~$\LaTeX$ by ciceronii

Solution 2

Call the row x, and the number from the leftmost side t. Call the first term in the ratio $N$. $N = \dbinom{x}{t}$. The next term is $N * \frac{x-t}{t+1}$, and the final term is $N * \frac{(x-t)*(x-t-1)}{(t+1)*(t+2)}$. Because we have the ratio, $N : N * \frac{x-t}{t+1} : N * \frac{(x-t)*(x-t-1)}{(t+1)*(t+2)}$ = $3:4:5$,

$\frac{x-t}{t+1} = \frac{4}{3}$ and $\frac{(x-t)*(x-t-1)}{(t+1)*(t+2)} = \frac{5}{3}$. Solve the equation to get get $t= 26$ and $x = \boxed{062}$.

https://www.wolframalpha.com/input/?i=%28x-t%29%2F%28t%2B1%29++%3D+4%2F3%2C++%28x-t%29%28x-t-1%29%2F%28%28t%2B1%29%28t%2B2%29%29+%3D+5%2F3


-Solution and LaTeX by jackshi2006


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. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1992_AIME_Problems/Problem_4&oldid=195846"