Question

Find the number of consecutive zeroes at the end of the given number.
$1!\times 2!\times 3!\times 4!\times 5!\times \cdots \cdots \times 50!$

• A. 235
• B. 12
• C. 262
• D. 105

Answer 1

The correct option is C 262

1! to 4! would have no zeroes while 5! to 9!, all the values would have 1 zero. Thus, a total of 5 zeroes till 9!. Going further 10! to 14! would have two zeroes each - so a total of 10 zeroes would come out of the product of $10!\times 11!\times 12!\times 13!\times 14!$
Continuing this line of thought further we get:
Number of zeroes between $15!\times 16!\cdots \times 19!$ = 3 + 3 + 3 + 3 + 3 = 3 $\times$ 5 = 15
Number of zeroes between $20!\times 21!\cdots \times 24!=4\times 5=20$
Number of zeroes between $25!\times 26!\cdots \times 29!=6\times 5=30$
Number of zeroes between $30!\times 31!\cdots \times 34!=7\times 5=35$
Number of zeroes between $35!\times 36!\cdots \times 39!=8\times 5=40$
Number of zeroes between $40!\times 41!\cdots \times 44!=9\times 5=45$
Number of zeroes between $45!\times 46!\cdots \times 49!=10\times 5=50$
Number of zeroes for 50! = 12
Thus, the total number of zeroes for the expression $1!\times 2!\times 3!\cdots \times 50!$ = 5 + 10 +15 + 20 + 30 + 35 + 40 + 45 + 50 + 12 = 262 zeroes. Option (c) is correct.