Number Theory | PRMO 2019 | Problem 3

Let \(x_1\) be a positive real number and for every integer $n\geq1$ let \(x_{n+1}=1+x_{1}x_{2}...x_{n-1}x_{n}\). If \(x_{5}=43\). what is the sum of digits of the largest prime factor of \(x_{6}\).

Here \(x_5=1+x_1x_2x_3x_4\) then \(x_1x_2x_3x_4=42\)

\(x_6=1+x_1x_2x_3x_4x_5\)=1+(42)(43)=1807=(13)(139)

Then largest prime factor=139 then sum of digits=13

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