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# Row of Pascal Triangle | AIME I, 1992 | Question 4

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1992 based on Row of Pascal Triangle.

## Row of Pascal Triangle - AIME I, 1992

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

• is 107
• is 62
• is 840
• cannot be determined from the given information

### Key Concepts

Integers

Digits

Combinatorics

AIME I, 1992, Question 4

Elementary Number Theory by David Burton

## Try with Hints

First hint

For consecutive entries

$\frac{{n \choose (x-1)}}{3}=\frac{{n \choose x}}{4}=\frac{{n \choose {x+1}}}{5}$

from first two terms $\frac{n!}{3(x-1)!(n-x+1)!}=\frac{n!}{4x!(n-x)!}$

$\Rightarrow \frac{1}{3(n-x+1)}=\frac{1}{4x}$

$\Rightarrow \frac{3(n+1)}{7}=x$ is first equation

Second Hint

for the next two terms

$\frac{n!}{4(n-x)!x!}=\frac{n!}{5(n-x-1)!(x+1)!}$

$\Rightarrow \frac{4(n-x)}{5}=x+1$

$\Rightarrow \frac{4n}{5}=\frac{9x}{5}+1$

Final Step

from first equation putting value of x here gives

$\Rightarrow \frac{4n}{5}=\frac{9 \times 3(n+1)}{5 \times 7}+1$

$\Rightarrow n=62, x=\frac{3(62+1)}{7}=27$

$\Rightarrow$ n=62.