I want to find the last two digits of the perfect number $2^{126}(2^{127}-1)$. I know that in order to do this we want to work modulo $100$:
$$2^{126}(2^{127}-1)\ (mod 100)$$
But it doesn't seem as though Fermat's little theorem or Euler's theorem are any use. Any ideas?
Best Answer
Let $$x=2^{126}(2^{127}-1)$$ we wish to calculate last two digits of $x$
since $4|x $ now we will try out to find $\; x\mod25$
using Euler's therorem $x\equiv 2^{6}(2^7-1) \mod25\equiv (-2^3-2^6) \mod25$
$x\equiv-22\mod25\equiv 3\mod25$
since $4|x\Rightarrow \exists \;k\in \mathbb{Z}$ such that $x=4k$ now $4k\equiv 3\mod 25$
$$k=3\times 4^{-1}\mod 25\equiv 7 \mod 25$$
Now $x=4(25q+7)=100q+28$
Hence last two digits are 28