Question 1179135

Assuming that 

{{{4*10^(2n)+9*10^(2n-1)+5=99k}}}

{{{4*10^(2n+1)+9*10^(2n+1)+5}}}

={{{100*4*10^(2n)+100*9*10^(2n-1)+5}}}

={{{100*4*10^(2n)+9*10^(2n-1)+5}}}

={{{100*(4*10^(2n)+9*10^(2n-1+5-5))+5}}}

={{{100*(4*10^(2n)+9*10^(2n-1+5))-500+5}}}

={{{100*99k-495}}}

={{{100*99k-5*99}}}

={{{(100k-5)*99}}} => {{{99}}} is a factor, means it is divisible by {{{99 }}}