BTGmoderatorDC wrote:If there are more than two numbers in a certain list, is each of the numbers in the list equal to 0?

(1) The product of any two numbers in the list is equal to 0.
(2) The sum of any two numbers in the list is equal to 0.
Source: GMAT Prep

\[L = \left\{ {\,{x_1}\,,\,{x_2}\,,\, \ldots \,\,,\,\,{x_n}} \right\}\,\,\,\,,\,\,\,n \geqslant 3\]
\[?\,\,\,:\,\,\,{\text{all}}\,\,{\text{zero}}\]
\[\left( 1 \right)\,\,\,{x_j} \cdot {x_k} = 0\,\,\,\,\,\left( {j \ne k} \right)\,\,\,\,\,\left\{ \begin{gathered}
\,{\text{Take}}\,\,L = \left\{ {0,0, \ldots ,0,0} \right\}\,\,\,\, \Rightarrow \,\,\,\left\langle {{\text{YES}}} \right\rangle \,\, \hfill \\
\,{\text{Take}}\,\,L = \left\{ {0,0, \ldots ,0,1} \right\}\,\,\,\, \Rightarrow \,\,\,\left\langle {{\text{NO}}} \right\rangle \,\,\, \hfill \\
\end{gathered} \right.\]

What about statement (2)? Do you "feel" this statement is sufficient... but you cannot be 100% sure?

EMBRACE MATHEMATICS and develop your quantitative maturity to EXCEL IN YOUR EXAM (and in the MBA that goes right after it)!

\[\left( 2 \right)\,\,\left\{ \begin{gathered}
\,{x_j} + {x_k} = 0 \hfill \\
{x_k} + {x_m} = 0 \hfill \\
\end{gathered} \right.\,\,\,\,\,\mathop \Rightarrow \limits^{\left( - \right)} \,\,\,\,\,\,{x_j} - {x_m} = 0\,\,\,\,\,\, \Rightarrow \,\,\,\,\,{x_j} = {x_m}\,\,\,\,{\text{for}}\,\,\,\underline {{\text{ANY}}} \,\,\,\,{x_j}\,,\,\,{x_k}\,,\,\,{x_m}\,\,\,{\text{in}}\,\,L\]
\[\,\left\{ \begin{gathered}
\,{x_j} = {x_m} \hfill \\
\,0 = {x_j} + {x_m} = 2\,\, \cdot {x_j} \hfill \\
\end{gathered} \right.\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,{x_j} = 0\,\,\,{\text{for}}\,\,{\text{all}}\,\,j\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,\,\left\langle {{\text{YES}}} \right\rangle \]

This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.

I was asked if there is another formal proof of the sufficiency of the statement (2) but in a more "down-to-earth" arguments.

Certainly!

Let´s imagine (at first) that there is a negative number among the elements in the given list, say A.
In this case, there is another number in the list (say B) such that A+B= 0, hence B must be positive (B=-A).
Let´s consider any third number (say C) of the list. (We know the list has at least three elements.)
It is impossible to have A+C = 0 (C would be positive) and B+C = 0 (C would be negative) simultaneously,
therefore there is NO negative number among the elements of the given list.

Let´s now imagine that there is a positive number among the elements in the given list, say B.
In this case, there is a negative number (say A) so that B+A = 0 (A=-B), but we have already proven
(in the previous paragraph) that there are NO negative elements in the given list.

From both paragraphs above, we are sure all numbers (elements) in the given list must be non-negative
and also non-positive, hence all of them are equal to zero.

Regards,
Fabio.