Finding all possible values of $\frac{a}{|a|} + \frac{b}{|b|} + \frac{c}{|c|} + \frac{abc}{|abc|}$

Question

Given that $a$, $b$, and $c$ are nonzero real numbers, find all possible values of the expression $\frac{a}{|a|} + \frac{b}{|b|} + \frac{c}{|c|} + \frac{abc}{|abc|}.$Enter all possible values, separated by commas.

I'm not sure what this problem wants. Aren't there infinite possible values to the expression?

Answer 1

Let's make cases

Case1 a,b,c>0 hence abc>0 therefore output is 4

Case2 a,b>0 c<0 hence abc<0 therefore output is 0

Case3 a>0 b,c<0 hence abc>0 therefore output is 0

Case4 a,b,c<0 hence abc<0 therefore output is -4

Hence the outputs of the given expression are {4,0,-4} and that's the answer.

Answer 2

Here is a generalisation. But for this problem it is much easier to find the values of the required expression directly. This answer merely provides a way to deal with a more general situation.

Let $a_1,a_2,\ldots,a_n\in\{-1,1\}$. Let $S_k$ denote the set of all subsets of size $k$ of $\{1,2,\ldots,n\}$. Fix a subset $A$ of $\{0,1,2,\ldots,n\}$. Define $$f_A(a_1,a_2,\ldots,a_n)=\sum_{k\in A}\sum_{T\in S_k}\prod_{j\in T}a_j.$$ (In the extreme situation where $A=\emptyset$, we have $f_A=0$. If $A=\{0,1,2,\ldots,n\}$, then $f_A(a_1,a_2,\ldots,a_n)=\prod_{j=1}^n(1+a_j)$.) Clearly, the maximum possible value of $f_A$ is $$f_A(1,1,\ldots,1)=\sum_{k\in A}\binom{n}{k}.$$

If $a_1=a_2=\ldots=a_r=-1$ and $a_{r+1}=a_{r+2}=\ldots=a_n=1$, then let $T_-=\{1,2,\ldots,r\}\cap T$ for each $T\subseteq \{1,2,\ldots,n\}$. Therefore, $$f_A(a_1,a_2,\ldots,a_n)=\sum_{k\in A}\sum_{p=0}^k\sum_{\substack{T\in S_k\\ |T_-|=p}}(-1)^p=\sum_{k\in A}\sum_{p=0}^k(-1)^p\binom{r}{p}\binom{n-r}{k-p}.$$ So all possible values of $f_A$ are $m_A(0)$, $m_A(1)$, $\ldots$, $m_A(n)$, where $$m_A(r)=\sum_{k\in A}\sum_{p=0}^k(-1)^p\binom{r}{p}\binom{n-r}{k-p}.$$

If $n=3$ and $A=\{1,3\}$, then $m_A(0)=3+1=4$, $m_A(1)=\big(2+(-1)\big)+(-1)=0$, $m_A(2)=\big((-2)+1\big)+1=0$, and $m_A(3)=(-3)+(-1)=-4$.