If 60! is written out as an integer, with how many consecutive 0's will that integer end?
a. 6
b. 12
c.14
d.42
e.56
c
a. 6
b. 12
c.14
d.42
e.56
c
60! = 1 * 2 * 3 * ..... * 60Viper83 wrote:If 60! is written out as an integer, with how many consecutive 0's will that integer end?
a. 6
b. 12
c.14
d.42
e.56
c
When do we get a zero at the end of a product?
When we multiple 10 to number we get a Zero at the end of the resulting product.
Thus, our main objective here is to find number of 10s staying between 1 and 60.
How many 10s are possible? To find this we need to find how many 5s are there in between 1 and 60. (Because every 5 under 60 when multipled by 2 creates a 10)
Number of 5s = 5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, 60 = 12.
But wait!! 25 and 50 have two fives each (25 = 5*5, 50 = 2*5*5). So these are to be added twice.
Thus number of possible 10s are 12 + 2 = 14
