Let S(M) denote the sum of the digits of a positive integer M written in base 10. Let N be the smallest positive integer such that S(N) = 2017. Find the value of S(5N + 2017)

Answer

To solve the problem, we need to find the smallest positive integer \( N \) such that the sum of its digits \( S(N) = 2017 \), and then calculate \( S(5N + 2017) \).

### Step 1: Determine \( N \)
The smallest positive integer \( N \) such that \( S(N) = 2017 \) can be constructed by using the digit '9' as much as possible.

- The maximum sum of digits we can achieve with '9's is \( 9 \times k \) where \( k \) is the number of '9's.
- To find \( k \), we solve \( 9k \leq 2017 \).

Calculating \( k \):
\[ k = \left\lfloor \frac{2017}{9} \right\rfloor = 223 \]
This gives us a sum of digits of:
\[ 9 \times 223 = 2007 \]

### Step 2: Construct \( N \)
To reach a sum of 2017, we need an additional \( 2017 - 2007 = 10 \).

Thus, we can construct \( N \) as:
- Use 223 '9's, and then add a '1' and a '0' to make the sum of the digits equal to 2017.

So, \( N = 999...9910 \) (223 '9's followed by '1' and '0').

### Step 3: Calculate \( 5N + 2017 \)
Now, we need to compute \( 5N + 2017 \):
\[ 5N = 5 \times (999...9910) \]
Calculating \( 5N \):
- The number \( N \) has 223 '9's, so \( 5N \) will have:
- \( 5 \times 9 = 45 \) for the first 223 digits, which results in carrying over.

This means:
- The first 222 digits will be '4' (due to carry) and the last digit will be '5' (from the last '10' in \( N \)).
- Thus, \( 5N \) will look like \( 444...4450 \) (222 '4's followed by '5' and '0').

Now, adding 2017:
\[ 5N + 2017 = 444...4450 + 2017 \]

### Step 4: Calculate the sum of digits \( S(5N + 2017) \)
Now we need to calculate the sum of the digits of \( 5N + 2017 \):
- The number \( 2017 \) has digits '2', '0', '1', '7'.

Adding \( 2017 \) to \( 444...4450 \):
- The last digits will be \( 0 + 7 = 7 \).
- The second last digits will be \( 5 + 1 = 6 \).
- The remaining digits will remain '4's.

Thus, \( 5N + 2017 \) will look like \( 444...4467 \) (222 '4's followed by '6' and '7').

### Final Calculation of \( S(5N + 2017) \):
Now, we calculate the sum of the digits:
\[ S(5N + 2017) = 222 \times 4 + 6 + 7 \]
Calculating:
\[ S(5N + 2017) = 888 + 6 + 7 = 901 \]

Thus, the final answer is:
\[ \boxed{901} \]