Bunuel wrote:

Official Solution:

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 the same as that of \((7^3)^4\) and the units digit of \(1973^{3^2}\) is the 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 (cyclisity): 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. \(7^2=49\) (last digit is 9)

3. \(7^3=xx3\) (last digit is 3)

4. \(7^4=xxx1\) (last digit is 1)

5. \(7^5=xxxxxx7\) (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, the units digit of \(7^{12}\) will be 1 (4th in pattern, as 12 is a multiple of cyclisty 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 than 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

Hi

BunuelCan you please explain how is 9 first in the cyclisty of number 3 ??

"the units digit of 3^9 will be 3 (first in pattern, as 9=4*2+1)"

Thanks