[Algebra]
Problem: A positive integer $n$ is called "savage" if the integers $\{1,2,\dots,n\}$ can be partitioned into three sets $A, B$ and $C$ such that

i) the sum of the elements in each of $A, B$, and $C$ is the same,

ii) $A$ contains only odd numbers,

iii) $B$ contains only even numbers, and

iv) C contains every multiple of 3 (and possibly other numbers).
Determine all even savage integers less than 100.

Solution: First, we prove lemma (b): if $n$ is an even savage integer, then $\frac{n+4}{12}$ is an integer.

Proof of lemma (b):
We use the strategy of putting all of the multiples of 3 between 1 and $n$ in the set $C$, all of the remaining even numbers in the set $B$, and all of the remaining numbers in the set $A$. The sums of these sets will not likely all be equal, but we then try to adjust the sums to by moving elements out of $A$ and $B$ into $C$ to try to make these sums equal. (Notice that we can't move elements either into $A$ or $B$, or out of $C$.) We will use the notation $|C|$ to denote the sum of the elements of $C$.

Since we are considering the case of $n$ even and we want to examine multiples of 3 less than or equal to $n$, it makes sense to consider $n$ as having one of the three forms $6 k$, $6 k+2$ or $6 k+4$. (These forms allow us to quickly tell what the greatest multiple of 3 less than $n$ is.)

Case 1: $n=6 k$

In this case, $C$ contains at least the integers $3,6,9, \ldots, 6 k$, and so the sum of $C$ is greater than one-third of the sum of the integers from 1 to $n$, since if we divide the integers from 1 to $n=6 k$ into groups of 3 consecutive integers starting with 1,2, 3 , then the set $C$ will always contain the largest of the 3 .



Case 2: $n=6 k+4$

Here, the sum of the integers from 1 to $n=6 k+4$ is $\frac{1}{2}(6 k+4)(6 k+5)=18 k^{2}+27 k+10=3\left(6 k^{2}+9 k+3\right)+1$, which is never divisible by 3 . Therefore, $n$ cannot be savage in this case because the integers from 1 to $n$ cannot be partitioned into 3 sets with equal sums.

Case 3: $n=6 k+2$

Here, the sum of the integers from 1 to $n=6 k+2$ is

$\frac{1}{2}(6 k+2)(6 k+3)=18 k^{2}+15 k+3$, so the sum of the elements of each of the sets $A, B$ and $C$ should be $6 k^{2}+5 k+1$, so that the sums are equal.

In this case $C$, contains at least the integers $3,6,9, \ldots, 6 k$, and so $|C| \geq 3+6+9+\cdots 6 k=3(1+2+3+\cdots+2 k)=3\left(\frac{1}{2}(2 k)(2 k+1)\right)=6 k^{2}+3 k$

The set $A$ contains at most the integers $1,3,5,7, \ldots, 6 k+1$, but does not contain the odd multiples of 3 less than $n$, ie. the integers $3,9,15, \ldots, 6 k-3$. Therefore, $|A| \leq(1+3+5+\cdots+6 k+1)-(3+9+\cdots+6 k-3)$

$=\frac{1}{2}(3 k+1)[1+6 k+1]-\frac{1}{2}(k)[3+6 k-3]$

$=(3 k+1)(3 k+1)-k(3 k)$

$=6 k^{2}+6 k+1$

(To compute the sum of each of these arithmetic sequences, we use the fact that the sum of an arithmetic sequence is equal to half of the number of terms times the sum of the first and last terms.)

The set $B$ contains at most the integers $2,4,6,8, \ldots, 6 k+2$, but does not contain the even multiples of 3 less than $n$, ie. the integers $6,12, \ldots, 6 k$. Therefore, $|B| \leq(2+4+6+\cdots+6 k+2)-(6+12+\cdots+6 k)$

$=\frac{1}{2}(3 k+1)[2+6 k+2]-\frac{1}{2}(k)[6+6 k]$

$=(3 k+1)(3 k+2)-k(3 k+3)$

$=6 k^{2}+6 k+2$

Thus, the set $C$ is $2 k+1$ short of the desired sum, while the set $A$ has a sum that is $k$ too big and the set $B$ has a sum that is $k+1$ too big.

So in order to correct this, we would like to move elements from $A$ adding to $k$, and elements from $B$ which add to $k+1$ all to set $C$.



Since we are assuming that $n$ is savage, then this is possible, which means that $k+1$ must be even since every element in $B$ is even, so the sum of any number of elements of $B$ is even.

Therefore, $k$ is odd, and so $k=2 l+1$ for some integer $l$, and so

$n=6(2 l+1)+2=12 l+8$, ie. $\frac{n+4}{12}$ is an integer.

Having examined all cases, we see that if $n$ is an even savage integer, then $\frac{n+4}{12}$ is an integer.


From the proof of (b) above, the only possible even savage integers less than 100 are those satisfying the condition that $\frac{n+4}{12}$ is an integer, ie. $8,20,32,44,56,68,80,92$. We already know that 8 is savage, so we examine the remaining 7 possibilities.

We make a table of the possibilities, using the notation from the proof of (b):

| $n$ | $k$ | Sum of elements <br> to remove from $A$ | Sum of elements <br> to remove from $B$ | Possible? |
| :---: | :---: | :---: | :---: | :---: |
| 20 | 3 | 3 | 4 | No - cannot remove a sum of 3 from <br> A. |
| 32 | 5 | 5 | 6 | Yes - remove 5 from $A, 2$ and 4 <br> from $B$ |
| 44 | 7 | 7 | 8 | Yes - remove 7 from $A, 8$ from $B$ |
| 56 | 9 | 9 | 10 | No - cannot remove a sum of 9 from <br> A. |
| 68 | 11 | 11 | 12 | Yes - remove 11 from $A, 4$ and 8 <br> from $B$ |
| 80 | 13 | 13 | 14 | Yes - remove 13 from $A, 14$ from $B$ |
| 92 | 15 | 15 | 16 | No - cannot remove a sum of 15 <br> from $A$ (since could only use $1,5,7$, <br> 11,13 ) |

Therefore, the only even savage integers less than 100 are 8, 32, 44, 68 and 80.

Final answer: 8,32,44,68,80