What is the remainder when 7^(548) is divided by 10?

1. Translate the problem requirements: We need to find what number is left over when \(7^{548}\) is divided by 10. Since dividing by 10 gives us the remainder as the last digit, we're really asking: what is the last digit of \(7^{548}\)?
2. Identify the cyclical pattern of last digits: Calculate the first few powers of 7 to observe how the last digit repeats in a predictable cycle
3. Determine the cycle length and position: Find how many different last digits appear before the pattern repeats, then figure out where 548 falls within this cycle
4. Apply the pattern to find the answer: Use the cyclical position to determine the last digit of \(7^{548}\)

Execution of Strategic Approach

1. Translate the problem requirements

Let's start by understanding what this question is really asking. When we divide any number by 10, the remainder is simply the last digit of that number. For example:

• 23 divided by 10 gives quotient 2 and remainder 3 (the last digit)
• 157 divided by 10 gives quotient 15 and remainder 7 (the last digit)
• 7548 divided by 10 gives quotient 754 and remainder 8 (the last digit)

So when we're asked "What is the remainder when \(7^{548}\) is divided by 10?", we're really asking: "What is the last digit of \(7^{548}\)?"

2. Identify the cyclical pattern of last digits

Now let's calculate the first several powers of 7 and focus only on their last digits to see if there's a pattern:

• \(7^1 = 7\) → last digit is 7
• \(7^2 = 49\) → last digit is 9
• \(7^3 = 7 \times 49 = 343\) → last digit is 3
• \(7^4 = 7 \times 343 = 2,401\) → last digit is 1
• \(7^5 = 7 \times 2,401 = 16,807\) → last digit is 7
• \(7^6 = 7 \times 16,807 = 117,649\) → last digit is 9

Look what happened! The last digits follow the pattern: 7, 9, 3, 1, 7, 9, ... The pattern is repeating every 4 powers.

This makes sense because to find the last digit of the next power, we only need to multiply the current last digit by 7 and take the last digit of that result:

• \(7 \times 7 = 49\) → last digit 9
• \(9 \times 7 = 63\) → last digit 3
• \(3 \times 7 = 21\) → last digit 1
• \(1 \times 7 = 7\) → last digit 7 (back to the start!)

3. Determine the cycle length and position

We've discovered that the last digits of powers of 7 repeat in a cycle of length 4:

Position 1 in cycle: last digit is 7 (like \(7^1, 7^5, 7^9\), ...)
Position 2 in cycle: last digit is 9 (like \(7^2, 7^6, 7^{10}\), ...)
Position 3 in cycle: last digit is 3 (like \(7^3, 7^7, 7^{11}\), ...)
Position 4 in cycle: last digit is 1 (like \(7^4, 7^8, 7^{12}\), ...)

To find where \(7^{548}\) falls in this cycle, we need to determine what position 548 corresponds to. We divide the exponent by the cycle length:

\(548 \div 4 = 137\) remainder 0

When the remainder is 0, this means we're at the end of a complete cycle, which corresponds to position 4 in our cycle.

4. Apply the pattern to find the answer

Since \(548 \div 4\) gives remainder 0, we know that \(7^{548}\) has the same last digit as \(7^4\).

From our pattern above, we found that \(7^4\) has last digit 1.

Therefore, \(7^{548}\) has last digit 1, which means the remainder when \(7^{548}\) is divided by 10 is 1.

Final Answer

The remainder when \(7^{548}\) is divided by 10 is 1.

This matches answer choice A.

Common Faltering Points

Errors while devising the approach

Faltering Point 1: Misunderstanding what "remainder when divided by 10" means
Many students may attempt to calculate \(7^{548}\) directly or use complex modular arithmetic formulas instead of recognizing that finding the remainder when any number is divided by 10 is simply finding its last digit. This leads to unnecessary complexity and potential calculation errors.

Faltering Point 2: Not recognizing the cyclical pattern approach
Students might try to compute higher powers of 7 manually (like \(7^5, 7^6\), etc.) without realizing they only need to track the last digits and look for a repeating pattern. This wastes time and increases the likelihood of arithmetic mistakes.

Errors while executing the approach

Faltering Point 1: Making arithmetic errors when computing the pattern
When calculating powers like \(7^2 = 49, 7^3 = 343\), or \(7^4 = 2,401\), students may make basic multiplication errors that would throw off the entire cyclical pattern. Even small mistakes here completely derail the solution.

Faltering Point 2: Incorrectly tracking only partial cycles
Students might stop after finding just 2-3 powers (like \(7^1 = 7, 7^2 = 9\)) and incorrectly conclude the pattern repeats every 2 steps, missing the complete 4-step cycle of 7, 9, 3, 1.

Errors while selecting the answer

Faltering Point 1: Misinterpreting remainder 0 in the division
When \(548 \div 4 = 137\) remainder 0, students often think this means position 0 in the cycle, but there is no position 0. A remainder of 0 actually corresponds to the last position in the cycle (position 4), which gives the last digit as 1, not 7.