+1926 We can use handy tricks to solve this problem.
Let's focus on the last digit of the sum. This will come in handy.
- 5 raised to any power will end in 5
7's last digit follows a pattern
\(7^1 = 7 \\ 7^2 = 49 \\ 7^3 = 343\\ 7^4 = 2401\)
- Since it repeats in groups of 4 7^13 ends in 7
\(5^19 + 7^13 + 23\) will end in \( 5 + 7 + 3 = 5 \)
This means 2 cannot be a divisor.
The smallest prime divisor wil either be 3 or 5
\([5^19 + 7^13 + 23] \pmod 3 = 2 \\ [5^19 + 7613 + 23 ] \pmod 5 = 0\)
The smallest prime divisor is 5
Thanks! :)
