Answer: Option C. -> 9
:
C
We need not consider ${2960}^{2960}$, as it will end in more number of zeroes.
Last non-zero digit of ${1390}^{2908}$ = Unit digit of ${139}^{2908}=1$.

Answer is (.......00 - .......1) = 9.

Answer: Option D. -> None of these
:
D
Consider first the number of 5-digit numbers divisible by 3. The smallest is $10002=3334\times 3$, the largest is $99999=33333\times 3$, so there are 30000 such numbers.
Now consider the numbers that do not contain the digit 6. There are 8 choices for the 1st digit, 9 choices for each of the 2nd, 3rd and 4th digits.
If the total of the first 4 digits is a multiple of 3, then the last digit must be 0, 3 or 9.
If it gives a remainder 1 when divided by 3, then the last digit must be 2, 5 or 8.
If it gives a remainder 2 when divided by 3, then the last digit must be 1, 4 or 7.
So in all cases there are 3 choices for the last digit. Hence the total number is $8\times 9^3\times 3=17496$.
So the total no. of 5 digit numbers which are divisible by 3 and contains a 6 is 12504.

Answer: Option D. -> None of these
:

The simplest way to find this is to put a rectangle around the given triangle and then subtract off the areas of the right triangles that surround it.
(See Figure)

The area of the entire rectangle is 126, but after subtracting the areas of the 3 right triangles, whose areas are 54, 3, and 42, we are left with an area of 27.

Answer: Option C. -> 72
:
C

The surface area of the solid consists of 6 faces of dimension 3 $\times$ 3 each with a 1 $\times$ 1 square hole.
The walls of the 6 holes can each be unfolded to form 1 $\times$ 4 rectangles.
Thus the surface area is 6(3$\times$3-1$\times$1)+6$\times$4 = 72