CGL 2012 Tier 2 Preparations and Results

Staff Selection Commission (SSC) SSC - CGL
2055

Q.How many zeroes are there in the product of natural numbers from 1 to 100?

2056
@sampras said: Q. What is the remainder of (n^n+n)/(n+1) where n is positive odd natural number.
A.(n-1) B.(n+1) C.n D.1
A.(n-1)
2057
@sumit1234567890 said: it is the time you ----home??
go, have gone, went, are going
go
2058
@gudda1122 said:
your answer is correct kindly post full method
Base of triangle is 80, n one of base angle is 60,the sum of length of othr two sides is 90.the
shortest side is?
kindly show full method
yr equate areas.......1/2*20*x*sin60 = Hero's formula
semi perimeter z (90+80)/2 = 85
2059
@sampras said: Q.How many zeroes are there in the product of natural numbers from 1 to 100?
24......
2060
@gudda1122 2
2061
@JS-M said:
24......
do we have any direct formula for this... like if its upto 1000 then..
2062
@sampras said: Q.How many zeroes are there in the product of natural numbers from 1 to 100?
100! mein zeros
=100/5+100/25=24
@sumit1234567890 said:
@gudda1122 4
bhai ek baar check kar lo shayd galat h

2063
@gudda1122
@gaurav0410
@sampras ans is-- went
2064
@sampras said:
do we have any direct formula for this... like if its upto 1000 then..
1000/5+1000/25+1000/125+1000/625
=200+40+8+1
=249
2065
@sampras said: Q.How many zeroes are there in the product of natural numbers from 1 to 100?
24....
2066
@gudda1122 tell me the ans of this i got 2
2067
@sampras said:
do we have any direct formula for this... like if its upto 1000 then..
yr u have to dividing d number by 5, 5^2, 5^3...........untill the number becomes smaller than the divisor
2068
@sumit1234567890 said:
@gudda1122 tell me the ans of this i got 2
bhai ur answer is correct
2069
@gudda1122 said:
if 1/(x+1) + 2/(y+2) + 1009/(z+1009) =1 then d value of x/(x+1) + y/(y+2) +z/(z+1009)
options r 0,2,3,4
kindly please tell me which value you took and why?
answer z 2...........put x=0, y=-4 and z=0
2070
@gudda1122 said:
100! mein zeros
=100/5+100/25=24
bhai ek baar check kar lo shayd galat h
yes its correct
@sampras said:
do we have any direct formula for this... like if its upto 1000 then..

@JS-M said:
24......
Oh got the formula.... the formula used is: (x/5)+ (x/5^2) + (x/5^3)+....

we have to keep increasing the power of 5 and adding the terms till 5^y is less than the last term of the multiplicand. Like 5^3=125 which will exceed 100 Hence only upto 5^2 is taken.
similarly for 1000 its 5^3 ... nearest being 625...
2071
@sampras said:
yes its correct
Oh got the formula.... the formula used is: (x/5)+ (x/5^2) + (x/5^3)+....we have to keep increasing the power of 5 and adding the terms till 5^x is less than the last term of the multiplicand. Like 5^3=125 which will exceed 100 Hence only upto 5^2 is taken.
similarly for 1000 its 5^3 ... nearest being 625...
yup bhai
isse hum kisi ki bhi kitni power hai yeh nikal sakte h
2072
@JS-M substarct -1 from each terma and add the same on the other side u will get the LHS same as that of question
2073
@gudda1122 said:
yup bhai
isse hum kisi ki bhi kitni power hai yeh nikal sakte h
yeah actually the logic behind using this formula is (10)^x= (2^m) * (5^n)

To find number of zeros i.e "x" lesser than "m" & "n" has to be taken...
2074

ONE WORD

Speech made without any preparation
ad-hoc, extempore, annals , fluke