x is the product of each integer from 1 to 50, inclusive and y=100^k, where k is an integer . What is the greatest value of k for which y is a factor of x?  Choose the correct answer.
A) 0
B) 5
C) 6
D) 10
E) 12
x = 1*2*3*4*5*......*49*50 = 50!   y = 100^k   100 = 10^2 = 2^2 * 5^2   On prime factorization of x, power of 5 will always be greater than power of 2   CONCEPT: In every Factorial value (x!), On prime factorization, the power of Bigger Prime number &gt; the power of Smaller Prime number   Also, Power of any Prime Number in any factorial can be calculated by following understanding   Power of prime x in n! = [n/x] + [n/x^2] + [n/x^3] + [n/x^4] + ... and so on   Where,   [x] = Greatest Integer less than or equal to x and the explanation of terms is as follows   [n/x] = No. of Integers that are multiple of x from 1 to n   [n/x^2] = No. of Integers that are multiple of x^2 from 1 to n whose first power has been counted in previous step and second is being counted at this step   [n/x^3] = No. of Integers that are multiple of x^3 from 1 to n whose first two powers have been counted in previous two step and third power is counted at this step   And so on.....   Where [n/x] is greatest Integer value of (n/x) less than or equal to (n/x)   i.e. [100/3] = [33.33] = 33   i.e. [100/9] = [11.11] = 11 etc.   Power of 5 in x = [50/5] + [50/5^2] + [50/5^3]... = 10 + 2 + 0= 
<gadget id="calculator">10 + 2 + 0</gadget>
<output>12</output>
12   Power of 2 in x= [50/2] + [50/2^2] + [50/2^3]... = 25 + 12 + 6 + 3 + 1= 
<gadget id="calculator">25 + 12 + 6 + 3 + 1</gadget>
<output>47</output>
47   i.e. x= 50! = 2^47 * 5^12 * ... = 10^12 * ... = 100^6 * ...   i.e. x is divisible by a 6th power of 100 at the most   i.e. y = 100^k = 100^6   i.e. k = 6   Answer: option 6
<result>C</result>
x is the product of each integer from 1 to 50, inclusive and y=100^k, where k is an integer . What is the greatest value of k for which y is a factor of x?