Question 4 :
options are :
0
1
2
3
more than 3
question 5:
OAs for the question at 3:10 pm
@saurav205 said:Question 4 :options are :0123more than 3
Options kya die h tumne?? Option 4) 1.07...Already been discussed
question 5:
OAs for the question at 3:10 pm
@ScareCrow28 said:Options kya die h tumne?? Option 4) 1.07...Already been discussed
Oh yes..Bhool gaya tha....I had already put up this question a few days back...
My bad...
@saurav205 said:One from my side :Options are :5060708090
is it 70??
let the numbers written by X,Y and Z be x,y and z resp...
y = 2x (hence y has both digits even)
z = y+10
=> 9
and y
and x has one odd and one even
x = 12, y =24 and z = 34
sum = 70...
@saurav205 said:Question 3 ;options are :9111215
even number ke 5's count karne hai...should be 12...
OAs:
70
Option 4
12
1.07(sorry this question was already discussed earlier and also the options were wrong)
2
My Take :
2> a E (-1 1) and b (-1 1)
3> no of zeros = 11 + 1 = 12 (realized the mistake 50 = 25*2)
@saurav205 said:OAs:70Option 4121.07(sorry this question was already discussed earlier and also the options were wrong)2
bhai 2 kaise?? dusra number kaunsa hogA?? 
the concept used is :
largest factor * smallest factor = second largest factor * second smallest factor * ..... = N
So ,
since N and 1 have been excluded :
let the second smallest factor be = S
and second largest be = L
L =21S
21S^2 = N
S can take values 2 and 3 (less than or equal to the smallest prime factor of 21)
hence 2 values.
@saurav205 said:OAs:70Option 4121.07(sorry this question was already discussed earlier and also the options were wrong)2
@Logrhythm said:bhai 2 kaise?? dusra number kaunsa hogA??
oh accha...2*(7*3*2) 
@Logrhythm said:21 = 7*3hence N = 3*3*7 so only 1??
bhai question theek se padho...you cannot include the number itself....
!> No of zeroes at the end of 25! +26! + 27! + 28! + 30!
2> No of zeroes at the end of 27!^27!