12. Let $N$ be the number of ordered sextuples of integers $(a, b, c, d, e, f)$ such that $a^{3}+b^{3}+c^{3}+d^{3}+e^{3}+f^{3}=168$ and $-202120212021^{9}<a b c d e f<202120212021^{9}$, where abcdef is the product of the six integers. What is the remainder of $N$ when divided by 6?
(A) 0
(B) 1
(C) 2
(D) 3
(E) 4

## Numerical Answer Problems -5 points
12. The answer is (C). Observe that if there exists a solution with $a<b<c<d<e<f$, then this contributes 6! ordered sextuples of solutions, so we can ignore it since we are counting the number of solutions modulo 6.

Similarly, for each partition $m_{1}+\cdots+m_{k}=6$ with $m_{i} \geq 1$, we can consider the solutions where there are $m_{i}$ coordinates equal to $x_{i}$, i.e., $\sum_{i=1}^{k} m_{i} x_{i}^{3}=168$. Each $k$-tuple of $x_{i}$ adds to the total number of ordered sextuples of solutions the quantity $\frac{6 !}{m_{1} ! \cdots m_{k} !}$.

We are therefore interested in the partitions for which the preceding number is not a multiple of 6; listing the partitions of 6 quickly, we see that the only ones we need to count are the solutions of type $(4,2),(3,3)$, and (6).

- Type (6), i.e., a single term with multiplicity 6; we then need to solve $6 a^{3}=168$, i.e., $a^{3}=28$, which has no solutions.
- Type (3,3); we then want to solve $3 a^{3}+3 b^{3}=168$, i.e., $a^{3}+b^{3}=56$. Factoring, we have $(a+b)\left(a^{2}-a b+b^{2}\right)=56$; for each signed divisor $d \mid 56$, we solve by substitution the system of second degree

$$
\left\{\begin{array}{l}
a+b=d \\
a^{2}-a b+b^{2}=\frac{56}{d}
\end{array}\right.
$$

The only solution is $(4,-2)$, so we add $\frac{6 !}{3 ! 3 !}=20 \equiv 2(\bmod 6)$ to the total.

- Type (4,2); we then want to solve $4 a^{3}+2 b^{3}=168$, i.e., $2 a^{3}+b^{3}=84$. We see that $b$ must be even, so let $b=2 c$, i.e., $2 a^{3}+8 c^{3}=84$; dividing by 2 gives $a^{3}+4 c^{3}=42$. Then $a$ must also be even, substituting $a=2 d$ we arrive at $8 d^{3}+4 c^{3}=42$ which is impossible because dividing by 2 would give $4 d^{3}+2 c^{3}=21$.

The answer is therefore 2, given by all permutations of the sextuple $(4,4,4,-2,-2,-2)$.