There are many ways in which the list $0,1,2,3,4,5,6,7,8,9$ can be separated into groups. For example, this list could be separated into the four groups $0,3,4,8$ and $1,2,7$ and 6 and 5,9. The sum of the numbers in each of these four groups is $15,10,6$, and 14 , respectively. In how many ways can the list $0,1,2,3,4,5,6,7,8,9$ be separated into at least two groups so that the sum of the numbers in each group is the same?
(A) 26
(B) 29
(C) 24
(D) 27
(E) 32
First, notice that the given list of 10 numbers has a sum of $0+1+2+3+4+5+6+7+8+9=45$. If $n$ is the number of groups and $m$ is the total in each group, then we must have $m n=45$. This means 45 is a multiple of the number of groups.

We also require that the number of groups be at least 2, so the number of groups is one of $3,5,9,15,45$.

If the number of groups is 9 , then the total in each group is $\frac{45}{9}=5$, but this is not possible since one of the groups must contain 9 and therefore cannot have a sum of 5 .

Similarly, if the number of groups is 15 or 45 , the sum of each group must be 3 or 1 respectively, which is too small.

Therefore, there are either 5 groups or 3 groups.

Let's first assume that there are five groups.

In this case, the total in each group must be $\frac{45}{5}=9$.

Since 0 does not contribute anything to the sum, we will ignore it for now.

Since the total in each group must be 9 , we are forced to have 9 in a group by itself.

The number 8 must be paired with 1 since adding any other positive integer to 8 will give a sum greater than 9 .

That is, one of the groups has the numbers 1 and 8 . We will denote this group by $\{1,8\}$.

The number 7 cannot be in a group with any number larger than 2 , and 1 is already paired with 8 , so another group must be $\{2,7\}$.

Continuing with this reasoning, we get that $\{3,6\}$ must be a group, and $\{4,5\}$ must also be a group.

Therefore, if there are 5 groups, they must be $\{9\},\{1,8\},\{2,7\},\{3,6\}$, and $\{4,5\}$.

As mentioned before, 0 does not contribute anything to the sum, so it can be placed in any of the 5 groups without changing the sum.

There are 5 ways to do this: $\{0,9\},\{1,8\},\{2,7\},\{3,6\},\{4,5\}$, or $\{9\},\{0,1,8\},\{2,7\},\{3,6\},\{4,5\}$, and so on.

We have shown that there are 5 ways to separate the numbers 0 through 9 into 5 groups so that each group has the same sum.

Let's now assume that there are three groups.

As we did when there were 5 groups, we ignore 0 for now.

Since there are three groups, the sum in each group is $\frac{45}{3}=15$.

We now find all groups which add up to 15 .

There are 17 of them, so we will label them using $A$ through $Q$ :

| $A$ | $\{6,9\}$ | $G$ | $\{2,5,8\}$ | $M$ | $\{1,3,4,7\}$ |
| :--- | :--- | :--- | :--- | :--- | :--- |
| $B$ | $\{1,5,9\}$ | $H$ | $\{3,4,8\}$ | $N$ | $\{4,5,6\}$ |
| $C$ | $\{2,4,9\}$ | $I$ | $\{1,2,4,8\}$ | $O$ | $\{1,3,5,6\}$ |
| $D$ | $\{1,2,3,9\}$ | $J$ | $\{2,6,7\}$ | $P$ | $\{2,3,4,6\}$ |
| $E$ | $\{7,8\}$ | $K$ | $\{3,5,7\}$ | $Q$ | $\{1,2,3,4,5\}$ |
| $F$ | $\{1,6,8\}$ | $L$ | $\{1,2,5,7\}$ |  |  |

Of these 17 groups, only $A, B, C$, and $D$ contain the number 9 .

Therefore, any way of separating the integers into three groups must use exactly one of these four groups.

Assume $A=\{6,9\}$ is one of the groups. This means the other two groups do not contain 6 or 9 .

The groups which lack both 6 and 9 are $E, G, H, I, K, L, M$, and $Q$.

If $E=\{7,8\}$ is one of the groups, we will have used $6,7,8$, and 9 , so the remaining group must be $Q=\{1,2,3,4,5\}$.

If $G=\{2,5,8\}$ is one of the groups, we will have used $2,5,6,8$, and 9 .

This forces the third group to be $M=\{1,3,4,7\}$.

Similarly, if $H$ is one of the groups, so is $L$, and if $I$ is one of the groups, so is $K$.

At this point, we can stop checking.

This is because, for example, if we assume $M$ is one of the groups, we are forced to take $G$ as the third group, but we have already identified the case where we separate into $A, M$, and $G$. To summarize, if $A$ is one of the groups, the possible configurations are

$$
A, E, Q \quad A, G, M \quad A, H, L \quad A, I, K
$$

If we assume $B$ is one of the groups, the other two groups must not contain any of 1,5 , and 9 . The groups satisfying this condition are $E, H, J$, and $P$.

By the same reasoning as in the previous paragraph, the configurations including $B$ are

$$
B, E, P \quad B, H, J
$$

The groups which have no members in common with $C$ are $E, F, K$, and $O$, and the configurations which use $C$ are

$$
C, E, O \quad C, F, K
$$

The only groups which have no members in common with $D$ are $E$ and $N$, so the only configuration using $D$ is

$$
D, E, N
$$

In total, we have found that there are 9 possible ways to separate the numbers from 1 through 9 into three groups each having the same sum.

As before, placing 0 in a group has no effect on the sum of that group. Since there are three groups, there are three possible ways to include 0 for each configuration, so there are $9 \times 3=27$ ways to separate the integers from 0 to 9 into three groups where each group has the same sum. Adding this to the five from earlier, the number of ways to separate the list $0,1,2,3,4,5,6,7,8,9$ into at least two groups so that the sum of the numbers in each group is the same is $27+5=32$.

ANsWer: (E)