[Ankurb]

Originally Posted by viveknitw

wat if the question ask 5000! ?
then this method is too handy.

yes, the method is good but I doubt whether we can use it while considering a factorial which is not a multiple of 10.
In your case
Z(15) = L(4^1)*Z(3)
you are neglecting digits 11,12,13,14
I guess, that's the error.

[meeramalekandat]

Originally Posted by shashank3012

a,a+d,a+2d,a+3d

5a=3a^2+14d^2+12ad

14d^2+12ad=5a-3a^2

d is negative,d^2 is positive

12ad=5a-3a^2-14d^2

d=-1=>3a^2-17a+14=0=>a=1 or a=-14/3

d=-2=>3a^2-29a+56=0=>a is a real number

d=-3=>3a^2-41a+126=0=>a is real number

d=-4=>3a^2-53a+224=0a is not real number.

so,3 possible values.

option (D)

i considered numbers as a-2d, a-d , a , a+d


hence , a-d +a + a+ d = 5 (a-2d)
i e 3a = 5a - 10d
ie 2a = 10d
a= 5d

I do not know how to proceed from here ....

can you help me solve in this way?

[Ankurb]

Originally Posted by meeramalekandat

Consider the sequence 1, -2, 3, -4, 5, -6………. whose nth term is [(-1)^(n+1)]n. Find the average of the first 200 terms of the sequence.

consider pairs
-1-1-1...-1(100 terms) = -100/200 = -1/2

[Ankurb]

Originally Posted by meeramalekandat

i considered numbers as a-2d, a-d , a , a+d


hence , a-d +a + a+ d = 5 (a-2d)
i e 3a = 5a - 10d
ie 2a = 10d
a= 5d

I do not know how to proceed from here ....

can you help me solve in this way?

Hey, question is squares of second, third and fourth terms of an A.P. is equal to 5 times its first term.

[ABhishek009]

Originally Posted by Ankurb

yes, the method is good but I doubt whether we can use it while considering a factorial which is not a multiple of 10.
In your case
Z(15) = L(4^1)*Z(3)
you are neglecting digits 11,12,13,14
I guess, that's the error.

I am totally ignorant about this method can anyone plz explain it ???

[uphorik_85]

Originally Posted by viveknitw

last non zero number of 15!
my approach-
so Z(15) = L(4^1)*Z(3)
z(3)=1*2*3=6
4^1=4
thus last digit 4

pls tell me where i m wrong!

to find the last non zero digit of any factorial remove all factors of 5 from it..
der ia funda dat..
1.the product of all numbers from 1to 10 not divisible by 5 ends in 6
i,e 1*2*3*4*6*7*8*9 =x6
2.product of all number from 11 to 20 not divisible by 5 ends in 6 i,e
11*12*13*14*16*17*18*19=x6
and so on...
now to get the rightmost non zero digit of a factorial we hav to divide it by the highest per of 5 and the same pwr of 2 ..
lets expand 15!
(15!)'=1*2*3*4*6*7*8*9 (complete)
*11*12*13*14
*1*2*3 (by removing factors of 5 i,e 5,10,15 becomes 1,2,3)
=6*4*6 (unit's digit)
=x4
hence the last digit of (15)'!=x4
so..far we hav divided 15! by 5^3..
now the quotient of division of (15!)' by 5^3 is x4 and is units digtis 4
now we hav to divide x4 by 2^3 tyo get the units digit of 15!
(15!)' %2^3 =R where Ris the rightmost nonzro digit of 15!
x4=8*xR
if R=3 then the units digit is forin the above equation

so the rightmost non zero digit of 15! is 3

[Ankurb]

Originally Posted by Decent_flirt

If f(x) = (2x²+21x-55)÷(2x²+5x+9) and g(x) = log (8-x)
4
How many roots does f(x)=g(x) have??
a)1
b)2
c)3
d)0
e)more than 3

anyone trying this??
I could find a range of 'x' in which the roots may exist but can't really find how many of them.
anyway, my take is option a)1

[|e0]

Originally Posted by uphorik_85

hi puys,
solve dis...
for what values of m does the equation
x^2 -4x -m -root(8^2 -32x -8m)
has exactly 2 distinct real solutions...

Problem reduces to comparing:
a)y=x^2 -4x -m ( parabola) ( parabola opening upwards)
b)y^2 =(8^2 -32x -8m) (only positive values of y)(parabola opening leftwards)

for m>0
a)positive x intercept = 2+sqrt(4+m) -1
b)positive x intercept = 64-8m/32 -2

for 2 distinct solutions (2) > (1)
i.e. 64-8m/32 > 2+sqrt(4+m) -(3)

solve it for values of m.

for m=0
2 distinct solution-(4)

for m<0
a)positive y intercept =-m
b)positive y intercept = 64-8m

for 2 distinct solutions 64-8m > -m -(5)
solve for values of m -


combine 3,4 and 5 to get value of m.

[Ankurb]

Originally Posted by uphorik_85

if R=3 then the units digit is forin the above equation

so the rightmost non zero digit of 15! is 3

one correction, R could be '8' too.
In fact,last non-zero digit of 15! is '8'

[ABhishek009]

Originally Posted by uphorik_85

to find the last non zero digit of any factorial remove all factors of 5 from it..
der ia funda dat..
1.the product of all numbers from 1to 10 not divisible by 5 ends in 6
i,e 1*2*3*4*6*7*8*9 =x6
2.product of all number from 11 to 20 not divisible by 5 ends in 6 i,e
11*12*13*14*16*17*18*19=x6
and so on...
now to get the rightmost non zero digit of a factorial we hav to divide it by the highest per of 5 and the same pwr of 2 ..
lets expand 15!
(15!)'=1*2*3*4*6*7*8*9 (complete)
*11*12*13*14
*1*2*3 (by removing factors of 5 i,e 5,10,15 becomes 1,2,3)
=6*4*6 (unit's digit)
=x4
hence the last digit of (15)'!=x4
so..far we hav divided 15! by 5^3..
now the quotient of division of (15!)' by 5^3 is x4 and is units digtis 4
now we hav to divide x4 by 2^3 tyo get the units digit of 15!
(15!)' %2^3 =R where Ris the rightmost nonzro digit of 15!
x4=8*xR
if R=3 then the units digit is forin the above equation

so the rightmost non zero digit of 15! is 3

LAst non zero digit must b 8 only.........

Consider last 3 digits only including zero

10 ! = 800
11! = 880
12! = 560
13! = 280
14 ! = 920
15 ! = 800


So last non zero digit must b 8