salsal wrote:

How many zeros does \(135^{19} \times 42^{17}\) end with?

Take the prime factorisations of the two numbers.

\(135 = 5 \times 3^3\\

42 = 2 \times 3 \times 7\)

\(135^{19} = 5^{19} \times {(3^3)}^{19}\\

42^{17} = 2^{17} \times 3^{17} \times 7^{17}\)

The prime factorisation of 10 is \(2 \times 5\)

\(135^{19} \times 42^{17} = \color{blue}{(2^{17} \times 5^{17})} \times 5^{2} \times {(3^3)}^{19} \times 3^{17} \times 7^{17} =\\

\color{blue}{10^{17}} \times 5^{2} \times {(3^3)}^{19} \times 3^{17} \times 7^{17}\)

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