Question 870998
{{{7^4*2*6^5+7^8*6^5=7^4*2*6^5+7^4*7^4*6^5=7^4*6^5*(2+7^4)=7^4*6^5*(2+2401)=7^4*6^5*2403}}}
Now we should "factorize" {{{6}}} and {{{2403}}}
{{{6=2*3}}}
{{{2403}}} is obviously divisible by {{{3}}} and {{{9}}} because its digits add up to {{{9}}} .
Dividing by {{{3}}} repeatedly we find that
{{{2403/3=801}}} , {{{801/3=267}}} , and {{{267/3=89}}} .
Since {{{89}}} is a prime number, the prime factorization of {{{3403is
{{{2403=3^3*89}}} .
So,
{{{7^4*2*6^5+7^8*6^5=7^4*6^5*2403=7^4*(2*3)^5*3^3*89=7^4*2^5*3^5*3^3*89=highlight(2^5*3^8*7^4*89)}}}