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.