Harsh9676 wrote:
Nups1324 wrote:
Harsh9676 wrote:
IMO A
we need to find the value of K for which 10^K and 12^K become factors of 50!
So 10 = 2*5 and 12 = 2^2 * 3
Minimum value of K should have two 2s, one 3s and one 5s.
We can infer that 50! has already two 2s and one 3. We just need to find the max power of 5 in 50!
Quotient of (50/5 )+ Quotient of (50/5^2) = 10 + 2 = 12
Ans. A
50! = 1*2*3*4*...*50. So 50! Has 2 twos and one 3.
We can also infer that the no of 2 s and no of 3 s in 50! Would be more than the no of 5s. This means that the no of 12s would be higher than no of 10s. So the limiting factor is 5.
So we need to find the no of 5 s in 50!.
Posted from my mobile deviceI'm really sorry, but how 50! has only two 2s and one 3.
1*2*3*(2*2)*5*(2*3).....*50?
This part I understood, "We can also infer that the no of 2 s and no of 3 s in 50! Would be more than the no of 5s. This means that the no of 12s would be higher than no of 10s. So the limiting factor is 5."
Secondly, what is this formula that you used to find the max power of 5 in 50! in your original answer.?
[ Quotient of (50/5 )+ Quotient of (50/5^2) = 10 + 2 = 12 ]
Thank you.