How many zeroes will be there at the end of $36!^{36!} ?$

7 \times 6!

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8 \times 6!

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7 \times 36!

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8 \times 36!

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Solution

The correct option is D $8 \times 36!$
Number of 5s in $36! = [\frac{36}{5}] + [\frac{36}{25}] = 7 + 1 = 8$
Therefore, zeros in $(36!)^{36!} = 8 \times 36!$

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