If n = 33^(43) + 43^(33) what is the units digit of n?

1. Translate the problem requirements: We need to find the units digit (the rightmost digit) of the sum \(33^{43} + 43^{33}\). This means we only care about what digit appears in the ones place of this large number.
2. Identify units digit patterns in powers: Find the repeating cycle of units digits when we raise numbers ending in 3 to successive powers, since both 33 and 43 end in 3.
3. Determine position within cycles: Use the exponents 43 and 33 to figure out where we are in each repeating cycle to find the specific units digits of \(33^{43}\) and \(43^{33}\).
4. Calculate the final units digit: Add the two units digits we found and determine the units digit of their sum.

Execution of Strategic Approach

1. Translate the problem requirements

We need to find the units digit of \(n = 33^{43} + 43^{33}\). The units digit is simply the rightmost digit of this very large number.

Think of it this way: if we calculated \(33^{43}\), we'd get some enormous number like 4,827,394,851,... something. We only care about that last digit. Same goes for \(43^{33}\). Then we add those two huge numbers together and want to know what the last digit of the sum is.

The key insight is that we don't actually need to calculate these massive numbers! We only need to track what happens to the units digit.

Process Skill: TRANSLATE - Converting the complex exponential expression into a simpler question about units digits

2. Identify units digit patterns in powers

Let's discover the pattern by looking at what happens when we raise numbers ending in 3 to different powers. Since both 33 and 43 end in 3, we only need to focus on powers of 3.

Let's calculate some powers of 3 and see what pattern emerges:

• \(3^1 = 3\) (units digit: 3)
• \(3^2 = 9\) (units digit: 9)
• \(3^3 = 27\) (units digit: 7)
• \(3^4 = 81\) (units digit: 1)
• \(3^5 = 243\) (units digit: 3)
• \(3^6 = 729\) (units digit: 9)

Notice the pattern in the units digits: 3, 9, 7, 1, 3, 9, 7, 1, ...

The units digits repeat in a cycle of 4: {3, 9, 7, 1}

This means that any number ending in 3, when raised to a power, will have a units digit that follows this same cycle. So \(33^n\) and \(43^n\) will have units digits that follow this exact pattern.

3. Determine position within cycles

Now we need to figure out where \(33^{43}\) and \(43^{33}\) fall within this 4-step cycle.

For \(33^{43}\): We need to find where the exponent 43 falls in our cycle of 4.
\(43 \div 4 = 10\) remainder \(3\)

This means \(43 = 4 \times 10 + 3\), so we go through the complete cycle 10 times, then move 3 more positions.
Position 3 in our cycle {3, 9, 7, 1} corresponds to 7.
Therefore, \(33^{43}\) has units digit 7.

For \(43^{33}\): We need to find where the exponent 33 falls in our cycle of 4.
\(33 \div 4 = 8\) remainder \(1\)

This means \(33 = 4 \times 8 + 1\), so we go through the complete cycle 8 times, then move 1 more position.
Position 1 in our cycle {3, 9, 7, 1} corresponds to 3.
Therefore, \(43^{33}\) has units digit 3.

Process Skill: INFER - Using remainder division to locate positions within repeating cycles

4. Calculate the final units digit

Now we add the units digits we found:

• \(33^{43}\) has units digit 7
• \(43^{33}\) has units digit 3

The units digit of their sum is the units digit of \(7 + 3 = 10\).

Since 10 has units digit 0, our final answer is 0.

Therefore, \(n = 33^{43} + 43^{33}\) has units digit 0.

Final Answer

The units digit of \(n\) is 0, which corresponds to answer choice A.

Verification: We found that \(33^{43}\) ends in 7, \(43^{33}\) ends in 3, and \(7 + 3 = 10\), which ends in 0.

Common Faltering Points

Errors while devising the approach

1. Attempting to calculate the actual values instead of using units digit patterns
Many students see \(33^{43} + 43^{33}\) and feel they need to compute these enormous numbers directly or use a calculator. They don't recognize that for units digit problems, we only need to track the pattern of the last digit through powers, making the problem much simpler.

2. Failing to recognize that both numbers follow the same pattern
Students might try to find separate patterns for \(33^n\) and \(43^n\) without realizing that since both base numbers end in 3, they follow the identical units digit cycle. This leads to unnecessary work and potential confusion.

3. Not understanding that units digits in addition work independently
Some students think they need to consider the full numbers when adding, rather than understanding that the units digit of a sum depends only on the units digits of the individual terms being added.

Errors while executing the approach

1. Making errors in the cyclical pattern calculation
When computing \(3^1, 3^2, 3^3, 3^4\) to find the pattern {3,9,7,1}, students often make arithmetic mistakes (like getting \(3^3 = 24\) instead of 27) or misread the units digits, leading to an incorrect cycle.

2. Confusing remainder positions with cycle positions
When dividing 43 by 4 to get remainder 3, students might think this means position 3 corresponds to the 3rd element in {3,9,7,1}, which is 7. However, they should remember that remainder 1 corresponds to position 1 (first element), remainder 2 to position 2, etc. A remainder of 0 corresponds to position 4 (last element).

3. Incorrect division or remainder calculation
Students may make basic arithmetic errors when calculating \(43 \div 4 = 10\) remainder \(3\), or \(33 \div 4 = 8\) remainder \(1\). These errors cascade through the rest of the solution.

4. Reporting the sum instead of the units digit of the sum
After correctly finding that one number ends in 7 and the other ends in 3, students calculate \(7 + 3 = 10\) but then select 10 as their answer, forgetting that the question asks specifically for the units digit of the sum.