Find the number of consecutive zeroes at the end of the given number.
100! + 200!

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Solution

The correct option is B 24
The number of zeroes would depend on the power of 5's in the value of the factorial.
100! would end in 20 + 4 = 24 zeroes
200! Would end in 40 + 8 + 1 = 49 zeroes.
When you add the two numbers (one with 24 zeroes and the other with 49 zeroes at it's end), the resultant total would end in 24 zeroes. Option (b) is correct.

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