Consider the following Qoutient: 12 11 * 10 * 9 * 8 * 7 *6 * 5 * 4 * 3 * 2 / 2^x (its 12 times 11 times 10 and so on... and "x" is in mid air after 2)
What is the maximum value of x such that 2^x is divisible into the product shown? (Hint: How many two's can divide evenly into the product shown in the numerator? Start "cancelling" two's)

Question

anonymous · Answer

...its not that hard guys..

anonymous · Answer

12 = 2*2*3  (running count = 2)
11 = prime
10 = 2*5  (3)
9 = not divisible by 2
8 = 2*2*2  (6)
7 = prime
6 = 2*3 (7)
5 = prime
4 = 2*2 (9)
3 = prime
2 = 2 (10)

x = 10?