61.
The number of zeros at the end of \((2!)^{2!} \times (5!)^{5!}\) is:
Answer: A
We evaluate the exponents first. \(2!=2\) and \(5!=120\). The expression is \((2!)^2 \times (5!)^{120}\). We need to count factors of 2 and 5. \(2! = 2\). \(5! = 120 = 2^3 \times 3 \times 5\). So the expression is \(2^2 \times (2^3 \times 3 \times 5)^{120} = 2^2 \times 2^{360} \times 3^{120} \times 5^{120} = 2^{362} \times 3^{120} \times 5^{120}\). We have 362 factors of 2 and 120 factors of 5. The number of pairs is limited by the smaller number, which is 120. So there are 120 trailing zeros.
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