M26-02

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 \(c^n\).

II. \({(a^m)}^n=a^{mn}\) and \(a^{m^n}=a^{(m^n)}\).

III. The units digit of integers in positive integer powers repeats in a specific pattern called cyclicity. The units digit of 7 and 3 in positive integer powers repeats in patterns of 4:

Hence, the units digit of \(17^{12}\) is 1, as the 12th number in the repeating pattern 7-9-3-1 is 1, and the units digit of \(1973^9\) is 3, as the 9th number in the repeating pattern 3-9-7-1 is 3.

BACK TO THE QUESTION:

Thus, we know that the units digit of \({(17^3)}^4=17^{12}\) is 1, and the units digit of \(1973^{3^2}=1973^9\) is 3. If the first number is greater than the second number, the units digit of their difference will be 8, for example, 11 - 3 = 8. However, if the second number is greater than the first number, the units digit of their difference will be 2, for example, 11 - 13 = -2.

Consider this, even if the first number were \(100^{12}\) instead of \(17^{12}\), and the second number were \(1000^9\) instead of \(1973^9\), the first number, \(100^{12}=10^{24}\), would still be smaller than the second number \(1,000^9=10^{27}\). Therefore, \(17^{12} < 1973^9\), and the units digit of the difference is 2.


Answer: B