What is the greatest integer k such that 2k is a factor of 67! ?

67!/2 = k = 18 235 555 459 094 342 644 124 929 548 302 732 213 583 817 657 024 762 296 850 814 250 133 981 218 471 936 000 000 000 000 000

about 1.82×10⁹⁴

k=64

Step-by-step explanation:

To find out the greatest integer such that 2^{k}is a factor if 67!

We divide the quotients by 2 and then finally add all the quotients.

67/2 = 33.5 > 33

33/2 = 16.5 > 16

16/2 = 8 > 8

8/2= 4 > 4

4/2  = 2> 2

2/2 = 1 > 1

k= 33+16+8+4+2+1

   = 64

Hence, 64 is the greatest integer such that 2^{k} is a factor of 67!.

false

step-by-step explanation:

i looked it up