joe writes N distinct positive integers on a board and calculates their pairwise sums. if all the digits from 0 to 9 appear in at least one of the units digits of the sums, find the smallest possible value of N

We are tasked with finding the smallest possible value of \( N \) such that Joe writes \( N \) distinct positive integers on a board and all digits from 0 to 9 appear as the units digits of the pairwise sums of these integers.

### Understanding the problem

Given \( N \) distinct integers \( x_1, x_2, \dots, x_N \), Joe calculates all their pairwise sums, which means he computes \( x_i + x_j \) for all \( 1 \leq i < j \leq N \). The challenge is to ensure that every digit from 0 to 9 appears at least once as the units digit in one of these sums.

### Key observations

- The units digit of a sum \( x_i + x_j \) depends only on the units digits of \( x_i \) and \( x_j \). For example, if \( x_i = 7 \) and \( x_j = 8 \), the sum \( x_i + x_j = 15 \), and its units digit is \( 5 \).
- Therefore, the problem reduces to selecting \( N \) distinct integers such that the set of all possible units digits of their pairwise sums covers all digits from 0 to 9.
- Since we are interested only in the units digits of the integers, without loss of generality, we can assume \( x_1, x_2, \dots, x_N \) have distinct units digits. That is, we can think of \( x_1, x_2, \dots, x_N \) as being distinct elements from the set \( \{0, 1, 2, \dots, 9\} \).

### Step-by-step approach

We want to select a set of distinct units digits such that the pairwise sums of these digits cover all digits from 0 to 9. Let’s examine how many distinct integers we need to achieve this.

1. If \( N = 2 \), the pairwise sums consist of only one sum, which gives only one possible units digit, so \( N = 2 \) is not enough.
   
2. If \( N = 3 \), the possible sums are \( x_1 + x_2 \), \( x_1 + x_3 \), and \( x_2 + x_3 \). This gives us at most 3 distinct units digits, so \( N = 3 \) is also not enough.

3. If \( N = 4 \), the possible sums are \( x_1 + x_2 \), \( x_1 + x_3 \), \( x_1 + x_4 \), \( x_2 + x_3 \), \( x_2 + x_4 \), and \( x_3 + x_4 \). This gives us 6 distinct sums, which is still insufficient to cover all digits from 0 to 9.

4. If \( N = 5 \), there are \( \binom{5}{2} = 10 \) possible pairwise sums. This means we can potentially have up to 10 distinct units digits, which could be enough to cover all digits from 0 to 9.

### Verifying with \( N = 5 \)

We now need to check if we can choose 5 distinct integers such that their pairwise sums yield all digits from 0 to 9 as units digits. A possible choice of integers is:

\[
\{0, 1, 2, 3, 9\}
\]

Let’s calculate the pairwise sums:

- \( 0 + 1 = 1 \)
- \( 0 + 2 = 2 \)
- \( 0 + 3 = 3 \)
- \( 0 + 9 = 9 \)
- \( 1 + 2 = 3 \)
- \( 1 + 3 = 4 \)
- \( 1 + 9 = 10 \) (units digit = 0)
- \( 2 + 3 = 5 \)
- \( 2 + 9 = 11 \) (units digit = 1)
- \( 3 + 9 = 12 \) (units digit = 2)

The distinct units digits we obtain are:

\[
\{0, 1, 2, 3, 4, 5, 9\}
\]

This set is still missing the digits 6, 7, and 8, so this choice does not work.

### Trying another set of integers

Let’s try the set:

\[
\{0, 1, 2, 4, 7\}
\]

The pairwise sums are:

- \( 0 + 1 = 1 \)
- \( 0 + 2 = 2 \)
- \( 0 + 4 = 4 \)
- \( 0 + 7 = 7 \)
- \( 1 + 2 = 3 \)
- \( 1 + 4 = 5 \)
- \( 1 + 7 = 8 \)
- \( 2 + 4 = 6 \)
- \( 2 + 7 = 9 \)
- \( 4 + 7 = 11 \) (units digit = 1)

The distinct units digits we obtain are:

\[
\{1, 2, 3, 4, 5, 6, 7, 8, 9\}
\]

This set still misses only the digit 0. Thus, adding the digit 3 (or changing one of the integers) might be needed. After adjustment, the smallest value comes out

Discover how to find the smallest N such that the pairwise sums of N distinct integers yield all digits from 0 to 9 as units digits. This problem involves concepts of number theory, modular arithmetic, and combinatorics, suitable for high school students in Grades 9-12.

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