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Bunuel
\(x_1+x_2+...+x_{100} = 1\);
\(x_1+x_2+...+x_{99} = 2\);
\(x_1+x_2+...+x_{98} = 3\);
...
\(x_1= 100\).

What is the value of \(x_1*x_2*...*x_{100} \)?

A. -100
B. -1
C. 0
D. 1
E. 100

Fresh GMAT Club Tests' Question


\(x_1+x_2+...+x_{100} = 1\);
\(x_1+x_2+...+x_{99} = 2\);
So \(2+x_{100}=1……x_{100}=-1\)

\(x_1+x_2+...+x_{99} = 2\);
\(x_1+x_2+...+x_{98} = 3\);
So, \(3+x_{98}=2………x_{98}=-1\)
...

and so on till \(x_2=1\)
Also, \(x_1= 100\).
Thus, \(x_1*x_2*...*x_{100}=100*-1*-1….*-1=100*(-1)^{99}=100*(-1)=-100 \)



A


I think there is a typo, it should be X2 = -1, right?
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\(x_1+x_2+...+x_{100} = 1\);
\(x_1+x_2+...+x_{99} = 2\);
\(x_1+x_2+...+x_{98} = 3\);
...
\(x_1= 100\).

What is the value of \(x_1*x_2*...*x_{100} \)?

A. -100
B. -1
C. 0
D. 1
E. 100

Fresh GMAT Club Tests' Question

Official Solution:


\(x_1+x_2+...+x_{100} = 1\);

\(x_1+x_2+...+x_{99} = 2\);

\(x_1+x_2+...+x_{98} = 3\);

\(...\)

\(x_1= 100\).

What is the value of \(x_1*x_2*...*x_{100}\)?


A. \(-100\)
B. \(-1\)
C. \(0\)
D. \(1\)
E. \(100\)


By subtracting the second equation from the first, we obtain \((x_1+x_2+...+x_{100})-(x_1+x_2+...+x_{99}) = 1-2\), which simplifies to \(x_{100}= -1\).

Similarly, subtracting the third equation from the second yields \((x_1+x_2+...+x_{99}) -(x_1+x_2+...+x_{98}) = 2-3\), which simplifies to \(x_{99}= -1\).

By repeating this process for each successive pair of equations, we can determine that every term up to \(x_1\) is also equal to -1.

Therefore, \(x_1*x_2*...*x_{100}=100(-1)(-1)(-1)...(-1)=100(-1)^{99}=-100\).


Answer: A
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