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

## Check the Answer

Answer: is 62.

AIME I, 1992, Question 4

Elementary Number Theory by David Burton

## Try with Hints

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

for the next two terms

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

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

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

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.

## Other useful links

- https://www.cheenta.com/rational-number-and-integer-prmo-2019-question-9/
- https://www.youtube.com/watch?v=lBPFR9xequA