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