24.33.17^2.24^2x.33^x.17^x.9^2x

24.33.17^2.(24^2.33.17.81)^x

Unit digit of 24.33.17^2=8

Unit digit of (24^2.33.17.81)^x=6[x can be any integer]

8*6=8

Pick D

(3*2^3)^(2x+1)*(3*11)^(x+1)*(17)^(x+2)*(3)^4x
(3)^(2x+1+x+1+4x)*2^(6x+3)*(11)^(x+1)*(17)^(x+2)
(3)^(7x+2)*2^(6x+3)*(11)^(x+1)*(17)^(x+2)

2,3,7 has cyclicity of 4
if x=1
3^9*2^9*11^2*17*3=> remainder=3*2*1*3=18
if x=2
3^16*2^15*11^3*17*4=> remainder=1*8*1*1=8
similarly for other x value we will get unit digit as 8
D

Hi All,

Since we have an equation given,

can we substitute a value for X.Let x=0

then the unit digit values would be **4 x **3 X **9 X 1 (as 9^0=1)

Then the answer is ****8.

Thanks
karthik

iikarthik wrote:Hi All,

Since we have an equation given,

can we substitute a value for X.Let x=0

then the unit digit values would be **4 x **3 X **9 X 1 (as 9^0=1)

Then the answer is ****8.

Thanks
karthik

The plug-in method is pretty fast in such problems.

However, we can not use zero as X since X has to be a positive integer.

My method is to plug in a integer for X

Let X = 1
Eqn becomes : 24^3 .33^2.17^3.9^2
:4*9*3*1
: __8

Pick D

pharmxanthan wrote:

iikarthik wrote:Hi All,

Since we have an equation given,

can we substitute a value for X.Let x=0

then the unit digit values would be **4 x **3 X **9 X 1 (as 9^0=1)

Then the answer is ****8.

Thanks
karthik

The plug-in method is pretty fast in such problems.

However, we can not use zero as X since X has to be a positive integer.

Hi

Thanks for correcting my mistake

gmatmachoman wrote:My method is to plug in a integer for X

Let X = 1
Eqn becomes : 24^3 .33^2.17^3.9^2
:4*9*3*1
: __8

Pick D

I too did this prob in the same way. I think this is the best approach to save time.