Question

Which of the following could be the units digit of \(57^n\), where n is a positive integer?
Indicate {all such digits.}

• A. 0
• B. 1
• C. 2
• D. 3
• E. 4

Hint:

To find the cyclicity of units digits, you only need to multiply the previous units digit by the base's units digit. For powers of 7: Start with 7. Then \(7 \times 7 = 49 \to 9\). Then \(9 \times 7 = 63 \to 3\). Then \(3 \times 7 = 21 \to 1\). Then \(1 \times 7 = 7\), and the pattern repeats.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
The units digit of a product of integers is determined solely by the units digits of the integers being multiplied. To find the possible units digits of \(57^n\), we only need to look at the pattern of the units digits of powers of 7. This pattern is cyclical.
Step 2: Key Approach:
We will calculate the first few powers of 7 and observe the pattern of their units digits. The pattern will repeat, and the set of digits in one cycle will be the set of all possible units digits.
Step 3: Detailed Explanation:
Let's find the units digits for the first few powers of 7.
- For n = 1: \(57^1\) ends in the same digit as \(7^1\), which is 7.
- For n = 2: \(57^2\) ends in the same digit as \(7^2 = 49\), which is 9.
- For n = 3: \(57^3\) ends in the same digit as \(7^3 = 7 \times 49\). The units digit is \(7 \times 9 = 63\), which is 3.
- For n = 4: \(57^4\) ends in the same digit as \(7^4 = 7 \times 343\). The units digit is \(7 \times 3 = 21\), which is 1.
- For n = 5: \(57^5\) ends in the same digit as \(7^5 = 7 \times 2401\). The units digit is \(7 \times 1 = 7\), which is 7.
The pattern of the units digits is 7, 9, 3, 1. This four-digit pattern repeats indefinitely. Therefore, the only possible units digits for any positive integer power of 57 are 1, 3, 7, and 9.
Step 4: Final Answer:
We check the given options to see which of them are in our set of possible digits \{1, 3, 7, 9\}.
The correct options are (B) 1, (D) 3, (H) 7, and (J) 9.