How many zeroes are in the solution of 65^20-65^16
\(\begin{array}{|rcll|} \hline && 65^{20}-65^{16} \\ &=& 65^{16}(65^4-1) \\ &=& 65^{16}(65^2-1)(65^2+1) \\ &=& (5\cdot 13)^{16}(4224)(4226) \quad \text{Decomposition into prime factors} \\ &=& (5\cdot 13)^{16}(2^7\cdot 3 \cdot 11)(2\cdot 2113 ) \quad 10~\text{is the factor of $5$ and $2$ } \\ &=& 2^8\cdot 5^{16}\ldots \quad \text{The minimum of $16$ and $8$ is $8$ is the number of zeroes in the product. } \\ \hline \end{array} \)
In the solution of \(\mathbf{65^{20}-65^{16}}\) are \(\mathbf{8}\) zeroes.
Proof \(65^{20}-65^{16} = 1\ 812\ 454\ 482\ 098\ 982\ 309\ 886\ 289\ 062\ 500\ 000\ 000\)