If m is an integer and m = 10^(32) - 32, what is the sum of the digits of m?

1. Translate the problem requirements: We need to find \(\mathrm{m} = 10^{32} - 32\), then calculate the sum of all digits in this resulting number.
2. Visualize the subtraction pattern: Understand what happens when we subtract 32 from a number that consists of 1 followed by 32 zeros.
3. Apply borrowing logic systematically: Work through the borrowing process that occurs when subtracting 32 from \(10^{32}\).
4. Count digits and calculate sum: Identify the pattern of digits in the result and sum them efficiently.

Execution of Strategic Approach

1. Translate the problem requirements

Let's break down what we're being asked to find. We have \(\mathrm{m} = 10^{32} - 32\), and we need to find the sum of all the digits in this number m.

First, let's understand what \(10^{32}\) looks like. This is the number 1 followed by 32 zeros:
\(10^{32} = 1000000000000000000000000000000000\) (32 zeros after the 1)

Now we need to subtract 32 from this number. The key insight is understanding what happens when we perform this subtraction.

Process Skill: TRANSLATE - Converting the problem statement into a clear mathematical understanding of what we need to calculate

2. Visualize the subtraction pattern

Let's think about this subtraction in everyday terms. Imagine you have 1 dollar (represented as 100 cents) and you need to subtract 32 cents. You'd get 68 cents.

But our situation is different - we're subtracting 32 from a number that looks like 1 followed by many zeros. Let's start with smaller examples to see the pattern:

- \(10^2 - 32 = 100 - 32 = 68\)
- \(10^3 - 32 = 1000 - 32 = 968\)
- \(10^4 - 32 = 10000 - 32 = 9968\)
- \(10^5 - 32 = 100000 - 32 = 99968\)

Do you see the pattern emerging? When we subtract 32 from \(10^n\) where \(n \geq 2\), we get a number that ends in 68, and all the digits before that are 9s.

Process Skill: VISUALIZE - Using concrete examples to see the pattern before applying it to the larger problem

3. Apply borrowing logic systematically

Now let's think about what happens when we subtract 32 from \(10^{32}\).

When we do this subtraction, we need to "borrow" from the 1 at the front. Here's how the borrowing works:

- The 1 becomes 0
- All the zeros in between become 9s (because of the borrowing chain)
- The last two digits become 68 (since \(100 - 32 = 68\))

So \(10^{32} - 32\) looks like this:
999999999999999999999999999999968

Let me count this out:

- We started with 1 followed by 32 zeros
- After subtraction, we have 30 nines, followed by 68
- Total digits: 30 nines + 2 digits (6 and 8) = 32 digits

4. Count digits and calculate sum

Now we can calculate the sum of all digits:

Our number has:

- Thirty 9s: \(30 \times 9 = 270\)
- One 6: 6
- One 8: 8

Sum of all digits = \(270 + 6 + 8 = 284\)

Let me verify this makes sense: We have 32 total digits, which matches our original \(10^{32}\) having 33 digits (1 followed by 32 zeros) becoming a 32-digit number after subtraction.

Final Answer

The sum of the digits of \(\mathrm{m} = 10^{32} - 32\) is 284.

This matches answer choice D.

We can verify our pattern is correct by checking that 30 + 2 = 32 digits total, and our sum calculation: \(30(9) + 6 + 8 = 270 + 14 = 284\).

Common Faltering Points

Errors while devising the approach

Students might confuse "sum of digits" with other mathematical operations. They might think they need to find the number of digits, or perform some other calculation on the number \(\mathrm{m} = 10^{32} - 32\), rather than adding up each individual digit that appears in the final number.

Students often attempt to calculate \(10^{32} - 32\) by trying to write out the full number or use standard arithmetic methods, rather than recognizing that there's a predictable pattern when subtracting from powers of 10. They might get overwhelmed by the large numbers and miss the systematic approach of using smaller examples to identify the pattern.

When students understand they need to use borrowing, they might incorrectly apply it. For example, they might think that only the last few digits change during subtraction, or they might not realize that the borrowing cascades through all the zeros, turning them into 9s.

Errors while executing the approach

After correctly determining that \(10^{32} - 32\) results in a string of 9s followed by 68, students often miscount how many 9s there are. Since \(10^{32}\) has 33 total digits (1 followed by 32 zeros), after subtraction we get 32 total digits. Students might incorrectly think there are 32 nines instead of 30 nines.

Even with the correct setup of 30 nines + 6 + 8, students can make simple arithmetic mistakes: calculating \(30 \times 9\) incorrectly (getting something other than 270), or making errors in the final addition \(270 + 6 + 8 = 284\).

When performing the subtraction for the rightmost digits, students might incorrectly calculate what \(100 - 32\) equals, or they might not properly understand how the borrowing affects just the last two positions, potentially getting digits other than 6 and 8.

Errors while selecting the answer

No likely faltering points - once students have calculated the sum correctly as 284, the answer selection is straightforward as it directly matches option D.