Let N = 1 + 11 + 111 + 1111 + ... + 111111111 (up to nine 1s). What is the sum of the digits of N?

Introduction / Context:
This question involves a sum of numbers made entirely of the digit 1 repeated different numbers of times, and then asks for the sum of the digits of the result. Such numbers are often called repunits. The problem tests your ability to handle pattern based sums and to work efficiently with large looking numbers by focusing on their structure rather than raw magnitude.

Given Data / Assumptions:

• The expression N is defined as N = 1 + 11 + 111 + 1111 + ... up to the term with nine 1s.
• The last term is 111111111 (nine 1s).
• There are nine terms in total in this sum.
• We must find the sum of the digits of the resulting number N.

Concept / Approach:
One direct approach is to recognize that adding 1, 11, 111, and so on in this pattern produces a very symmetric final number. In fact, if we add these terms carefully, we find that the result is 123456789. Once this is known or derived, the sum of the digits is easy to compute. Alternatively, we can simulate the process of adding these terms by looking at how many times each place value contributes to the final sum. However, for exam purposes, recognizing or verifying that N equals 123456789 is the fastest route.

Step-by-Step Solution:
Write out the nine terms explicitly: 1, 11, 111, 1111, 11111, 111111, 1111111, 11111111, and 111111111. Start adding them in a column format to observe the pattern in the sum. If you perform the addition carefully, you will see that carrying over digits creates a final number with digits increasing from 1 to 9. The final result of the sum is N = 123456789. Now find the sum of the digits of N: 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9. Compute this digit sum: 1 + 2 + 3 + 4 + 5 = 15 and 6 + 7 + 8 + 9 = 30. Add the two partial sums: 15 + 30 = 45. Therefore, the sum of the digits of N is 45.

Verification / Alternative Check:
To verify that N really is 123456789, you can add a few terms step by step. For example, 1 + 11 = 12. Then 12 + 111 = 123. Next 123 + 1111 = 1234. Continue this pattern and you will see the sum after four terms is 1234, after five terms is 12345, and so on, until after nine terms the sum is 123456789. This pattern arises because each new term ends with a 1 and creates a chain of carries that push the previous digits one place to the left while adding 1 to the new leading position.

Why Other Options Are Wrong:

• Option 18: This is much smaller than the true digit sum and may come from adding only a subset of the digits or miscalculating the final number.
• Option 36: This is the sum of 1 to 8 but ignores the ninth digit or results from an arithmetic slip.
• Option 5: This is far too small and suggests that only one or two digits were considered.
• Option 27: This is the sum of the digits of 999999999, not of 123456789, so it does not match this specific series.

Common Pitfalls:
The main difficulty is that the numbers look large and many students are tempted to give up without attempting the addition. Others may add quickly and make carrying mistakes, leading to an incorrect final value for N. Some may also mistakenly interpret the question and think they only need the number of terms or the last digit. Writing the addition vertically and checking carries carefully makes the process straightforward. Recognizing the well known pattern 1 + 11 + 111 + ... up to nine 1s yielding 123456789 is a useful shortcut.

Final Answer:
The sum of the digits of N is 45.