Use the binomial expansion of the expression $(10+1)^{100}$.

## Solution

$S=(10+1)^{100}-1=\ldots+(100 \cdot 99 \cdot 98: 6) \cdot 10^{3}+(100 \cdot 99: 2) \cdot 10^{2}+100 \cdot 10+1=\ldots+(33 \cdot 49) \cdot 10^{5}+495000+1000=A$ +496000, where all terms in place of the ellipsis are divisible by $10^{4}$, meaning $A$ ends with at least four zeros. Therefore, the number $S$ ends with three zeros (and the fourth last digit of the number $S$ is 6).

## Answer

Three zeros.
Find the number of zeros with which the number $11^{100}-1$ ends.

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