Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1992 based on Row of Pascal Triangle.
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.
Integers
Digits
Combinatorics
Answer: is 62.
AIME I, 1992, Question 4
Elementary Number Theory by David Burton
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.
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