2015/118) Without using logs and calculator find the number of digits in $2^{100}$

$2^{10} = 1024 > 1000 = 10^3$
so $2^{100} > 10^{30}$
further
$2^{10} < 1025 < 10^3 * \frac{41}{40}=  10^3 *( 1+ \frac{1}{40}) $
so $2^{100} < 10^{30} * ( 1+ \frac{1}{40})^{10} < 10^{30} * ( 1+ \frac{1}{40})^{40}<  10^{30} * e <  10^{30} * 3$
so $10^{30} < 2^{100} < 3 * 10^{30}$
hence 31 digits