Mountain14 wrote:

This is good question.

if the second number is smaller, then would it be 11-3 = 7?

It would be 11-3=8.

What is the units digit of (17^3)^4-1973^{3^2}?A. 0

B. 2

C. 4

D. 6

E. 8

Must know for the GMAT:

I. The units digit of

(abc)^n is the same as that of

c^n, which means that the units digit of

(17^3)^4 is that same as that of

(7^3)^4 and the units digit of

1973^{3^2} is that same as that of

3^{3^2}.

II. If exponentiation is indicated by stacked symbols, the rule is to work from the top down, thus:

a^m^n=a^{(m^n)} and not

(a^m)^n, which on the other hand equals to

a^{mn}.

So:

(a^m)^n=a^{mn};

a^m^n=a^{(m^n)}.

Thus,

(7^3)^4=7^{(3*4)}=7^{12} and

3^{3^2}=3^{(3^2)}=3^9.

III. The units digit of integers in positive integer power repeats in specific pattern (cyclicity): The units digit of 7 and 3 in positive integer power repeats in patterns of 4:

1. 7^1=7 (last digit is 7)

2. 7^2=9 (last digit is 9)

3. 7^3=3 (last digit is 3)

4. 7^4=1 (last digit is 1)5. 7^5=7 (last digit is 7 again!)

...

1. 3^1=3 (last digit is 3)

2. 3^2=9 (last digit is 9)

3. 3^3=27 (last digit is 7)

4. 3^4=81 (last digit is 1)5. 3^5=243 (last digit is 3 again!)

...

Thus th units digit of

7^{12} will be 1 (4th in pattern, as 12 is a multiple of cyclicty number 4) and the units digit of

3^9 will be 3 (first in pattern, as 9=4*2+

1).

So, we have that the units digit of

(17^3)^4=17^{12} is 1 and the units digit of

1973^3^2=1973^9 is 3. Also notice that

the second number is much larger then the first one, thus

their difference will be negative, something like 11-13=-2, which gives the final answer that the units digit of

(17^3)^4-1973^{3^2} is 2.

Answer B.