Question

The number of zeroes at the end of the sum ${101}^{11}-1$

Answer 1

$101\times 101=10201$

$(101{)}^2=1030301$

$(101{)}^4=104060401$

there is one pattern that in all the powers of $101$ last three digits

$M01$ where $M$ is a non zero digit.

So at the end of $(101{)}^n-1$

last three digits are $M00$

then there are $2$ zeroes at the end.