How many trailing zeroes would be found in 4617!, upon expansion?
Choose one:
A) 1150 B) 1151 C) 1152 D) 1153 E) 1154
for no of trailing zeros we have to find the no of 5 in 4617!,because 4617!=1*2*3*4*5*.........4616*4617.   no of 5 in 4617!= (4617/5^1)+(4617/5^2)+(4617/5^3)+(4617/5^4)+(4617/5^5)+... up to 5^n 4617.   now no of 5 =4617/5 + 4617/25 + 4617/125 + 4617/625 +4617/3125   =923 + 184 + 36 + 7 + 1 ( only take the integer value )   =1151   so the no of zeros at the end of 4617! is 1151.   ANSWER:1151
<result>B</result>
How many trailing zeroes would be found in 4617!, upon expansion?