rocko911 wrote:
chetan2u wrote:
Gnpth wrote:
Let N = 60! + 55! + 50! The unit digit of N and a number of digits to the left of the units digits are consecutive zeros before we come to the first non-zero digit. How many such consecutive zeros are there until the first non-zero digit?
A.16
B.15
C.14
D.12
E.10
Don't do the mistake of adding all separately..
Number of 0s will depend on number of 10s, which in turn would depend on 5s ...So \(N=60!+55!+50!=50!(51*52..60 + 51*52*..5 + 1)\)..
Don't calculate the bracket as the two terms are multiple of 5 but then 1 is added to it so entire bracket will NOT be div by 5 or 10So number of 5s in 50! =\(\frac{50}{5} + \frac{50}{5^2}=10+2=12\)
D
Posted from my mobile deviceHi, I have seen people doing this in many questions bt im still low at understanding it ....Please explain in most easiest way possible thank u
So number of 5s in 50! =\(\frac{50}{5} + \frac{50}{5^2}=10+2=12\)
hi...
LOGIC is taking OUT number of 5s that are in 50!
now \(50! =1*2*3*4*5*...49*50\)
MULTIPLES of 5 are 5,10,15,20,25,30,35,40,45,50
their product will have \(5*10*15*20*25*30*35*40*45*50\) ...... single 5 from each one of them so 50/5, this is nothing but multiples of 5 till 50
but their are multiples of \(5^2\), so an EXTRA 5, that is why \(\frac{50}{5^2}=2\) , these are 25 and 50
add all of them to get TOTAL..
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