snigdha1605 wrote:What is the units digit of (17)^12−(1973)^9?

0
2
4
6
8

OA = 2

When an integer is raised to consecutive powers, the resulting units digits repeat in a CYCLE.

17¹²:
7¹ --> units digit of 7.
7² --> units digit of 9. (Since the product of the preceding units digit and 7 = 7*7 = 49.)
7³ --> units digit of 3. (Since the product of the preceding units digit and 7 = 9*7 = 63.)
7� --> units digit of 1. (Since the product of the preceding units digit and 7 = 3*7 = 21.)
From here, the units digits will repeat in the same pattern: 7, 9, 3, 1.
The units digit repeat in a CYCLE OF 4.
Implication:
When an integer with a units digit of 7 is raised to a power that is a multiple of 4, the units digit will be 1.
Thus, 17¹² has a units digit of 1.

1973�:
3¹ --> units digit of 3.
3² --> units digit of 9. (Since the product of the preceding units digit and 3 = 3*3 = 9.)
3³ --> units digit of 7. (Since the product of the preceding units digit and 3 = 9*3 = 27.)
3� --> units digit of 1. (Since the product of the preceding units digit and 3 = 7*3 = 21.)
From here, the units digits will repeat in the same pattern: 3, 9, 7, 1.
The units digit repeat in a CYCLE OF 4.
Implication:
When an integer with a units digit of 3 is raised to a power that is a multiple of 4, the units digit will be 1.
Thus, 1973� has a units digit of 1.
From here, the cycle of units digits will repeat: 3, 9, 7, 1...
Thus, 1973� has a units digit of 3.

Result:
17¹² - 1973� = (integer with a units digit of 1) - (integer with a units digit of 3).
Tricky part: 17¹² < 1973�.
Thus, 17¹² - 1973� = (SMALLER integer with a units digit of 1) - (GREATER integer with a units digit of 3).
This is analogous to performing the following calculation:
11-23 = -12.
Thus, 17¹² - 1973� has a units digit of 2.

The correct answer is B.

To my knowledge, no official problem has ever asked for the units digit of a negative value.
Please feel free to ignore this problem.