Try this beautiful problem from Pre RMO, 2019 based on the Number theory.
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 \(x_{6}\).
Sequence
Series
Number System
But try the problem first...
Answer: is 13.
PRMO, 2019
Elementary Number Theory by David Burton
First hint
Here \(x_5=1+x_1x_2x_3x_4\) then \(x_1x_2x_3x_4=42\)
Second Hint
\(x_6=1+x_1x_2x_3x_4x_5\)=1+(42)(43)=1807=(13)(139)
Final Step
Then largest prime factor=139 then sum of digits=13
Try this beautiful problem from Pre RMO, 2019 based on the Number theory.
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 \(x_{6}\).
Sequence
Series
Number System
But try the problem first...
Answer: is 13.
PRMO, 2019
Elementary Number Theory by David Burton
First hint
Here \(x_5=1+x_1x_2x_3x_4\) then \(x_1x_2x_3x_4=42\)
Second Hint
\(x_6=1+x_1x_2x_3x_4x_5\)=1+(42)(43)=1807=(13)(139)
Final Step
Then largest prime factor=139 then sum of digits=13
