2022/23 AHSMC Part 1 Problem 11

Since $$\begin{align*} 25 &= 26 - 1 \\ &= 24 + 1 = 24 + 2 - 1 \\ &= 21 + 3 + 1 = 21 + 3 + 2 - 1 \\ &= 18 + 6 + 1 = 18 + 6 + 2 - 1 \\ &= 15 + 9 + 1 = 15 + 9 + 2 - 1 = 15 + 6 + 3 + 1 = 15 + 6 + 3 + 2 - 1 \\ &= 12 + 9 + 3 + 1 = 12 + 9 + 3 + 2 - 1 \end{align*}$$ we conclude that there are $13$ subsets whose elements sum to $25$. The answer is (E).

Alternative Solution: Let $S$ be a subset of $\{-1, 1, 2, 3, 6, 9, 12, 15, 18, 21, 24, 26\}$ whose elements sum to $25$. If $26 \in S$ then $S = \{-1, 26\}$. Next assume $26 \notin S$ and define $R = \{-1, 1, 2\}$ and $T = \{3, 6, 9, 12, 15, 18, 21, 24\}$ (note that $S \subseteq R \cup T$). Let $r$ be the sum of the elements in $S \setminus T$ and $t$ be the sum of the elements in $S \setminus R$. Note $-1 \le r \le 3$ since $S \setminus T \subseteq R$. Also note $t$ is divisible by $3$ since $S \setminus R \subseteq T$ and every number in $T$ is divisible by $3$. Since $r + t = 25$ it must be that $r = 1$ and $t = 24$. Therefore $S \setminus T$ is either $\{1\}$ or $\{-1, 2\}$, and $S \setminus R$ is one of the six subsets of $T$ whose elements sum to $24$: $\{24\}$, $\{3,21\}$, $\{6,18\}$, $\{9,15\}$, $\{3,6,15\}$ or $\{3,9,12\}$. Thus $S$ is one of the following twelve subsets: $\{1,24\}$, $\{1,3,21\}$, $\{1,6,18\}$, $\{1,9,15\}$, $\{1,3,6,15\}$, $\{1,3,9,12\}$, $\{-1,2,24\}$, $\{-1,2,3,21\}$, $\{-1,2,6,18\}$, $\{-1,2,9,15\}$, $\{-1,2,3,6,15\}$ or $\{-1,2,3,9,12\}$. Including $\{-1,26\}$, there are $13$ subsets whose elements sum to $25$.