Difference between revisions of "1992 AIME Problems/Problem 4"
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== Problem == | == 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 <math>3: 4: 5</math>? | + | 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. |
+ | |||
+ | <cmath>\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}</cmath> | ||
+ | In which row of [[Pascal's Triangle]] do three consecutive entries occur that are in the ratio <math>3 :4 :5</math>? | ||
== Solution 1== | == Solution 1== | ||
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==Solution 2== | ==Solution 2== | ||
− | Call the row x, and the number from the leftmost side t. Call the first term in the ratio <math>N</math>. <math>N = \dbinom{x}{t}</math>. The next term is <math>N * \frac{x-t}{t+1}</math>, and the final term is <math>N * \frac{(x-t)*(x-t-1)}{(t+1)*(t+2)}</math>. Because we have the ratio, <math>N : N * \frac{x-t}{t+1} : N * \frac{(x-t)*(x-t-1)}{(t+1)*(t+2)}</math> = <math>3:4:5</math> | + | Call the row x, and the number from the leftmost side t. Call the first term in the ratio <math>N</math>. <math>N = \dbinom{x}{t}</math>. The next term is <math>N * \frac{x-t}{t+1}</math>, and the final term is <math>N * \frac{(x-t)*(x-t-1)}{(t+1)*(t+2)}</math>. Because we have the ratio, <math>N : N * \frac{x-t}{t+1} : N * \frac{(x-t)*(x-t-1)}{(t+1)*(t+2)}</math> = <math>3:4:5</math>, |
<math>\frac{x-t}{t+1} = \frac{4}{3}</math> and <math>\frac{(x-t)*(x-t-1)}{(t+1)*(t+2)} = \frac{5}{3}</math>. | <math>\frac{x-t}{t+1} = \frac{4}{3}</math> and <math>\frac{(x-t)*(x-t-1)}{(t+1)*(t+2)} = \frac{5}{3}</math>. | ||
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− | -Solution and LaTeX by | + | -Solution and LaTeX by jackshi2006 |
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[[Category:Intermediate Number Theory Problems]] | [[Category:Intermediate Number Theory Problems]] | ||
{{MAA Notice}} | {{MAA Notice}} | ||
− |
Revision as of 00:11, 1 August 2021
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.
In which row of Pascal's Triangle do three consecutive entries occur that are in the ratio ?
Solution 1
Consider what the ratio means. Since we know that they are consecutive terms, we can say
Taking the first part, and using our expression for choose , Then, we can use the second part of the equation. Since we know we can plug this in, giving us We can also evaluate for , and find that Since we want , however, our final answer is ~LaTeX by ciceronii
Solution 2
Call the row x, and the number from the leftmost side t. Call the first term in the ratio . . The next term is , and the final term is . Because we have the ratio, = ,
and . Solve the equation to get get and .
-Solution and LaTeX by jackshi2006
1992 AIME (Problems • Answer Key • Resources) | ||
Preceded by Problem 3 |
Followed by Problem 5 | |
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All AIME Problems and Solutions |
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