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16 Sep 2014, 01:45
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Is the product of three consecutive integers negative?
(1) The sum of the integers is equal to the product of the integers.
(2) At least one of the integers is negative.
(1) The sum of the integers is equal to the product of the integers.
(2) At least one of the integers is negative.
16 Sep 2014, 01:45
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Official Solution:
Is the product of three consecutive integers negative?
(1) The sum of the integers is equal to the product of the integers.
Let the three consecutive integers be \((x-1)\), \(x\), and \((x+1)\). Hence, the above statement implies that \((x-1)+x+(x+1) = (x-1)x(x+1)\):
\(3x=x(x^2-1)\)
\(x(x^2-4)=0\)
Thus, \(x=-2\), \(x=0\), or \(x=2\).
The possible sets are: {-3, -2, -1}, {-1, 0, 1}, or {1, 2, 3}. Not sufficient.
(2) At least one of the integers is negative.
Considering this, we can have three scenarios:
(i) All three integers are negative. In this case, their product is negative.
(ii) Two of the integers are negative: {-2, -1, 0}. In this case the product will be zero.
(iii) Only one of the integers is negative: {-1, 0, 1}. In this case the product will be zero.
Not sufficient.
(1)+(2) We still can have 2 sets: {-3, -2, -1} and {-1, 0, 1}. So, the product can be negative as well as zero. Not sufficient.
Answer: E
Is the product of three consecutive integers negative?
(1) The sum of the integers is equal to the product of the integers.
Let the three consecutive integers be \((x-1)\), \(x\), and \((x+1)\). Hence, the above statement implies that \((x-1)+x+(x+1) = (x-1)x(x+1)\):
\(3x=x(x^2-1)\)
\(x(x^2-4)=0\)
Thus, \(x=-2\), \(x=0\), or \(x=2\).
The possible sets are: {-3, -2, -1}, {-1, 0, 1}, or {1, 2, 3}. Not sufficient.
(2) At least one of the integers is negative.
Considering this, we can have three scenarios:
(i) All three integers are negative. In this case, their product is negative.
(ii) Two of the integers are negative: {-2, -1, 0}. In this case the product will be zero.
(iii) Only one of the integers is negative: {-1, 0, 1}. In this case the product will be zero.
Not sufficient.
(1)+(2) We still can have 2 sets: {-3, -2, -1} and {-1, 0, 1}. So, the product can be negative as well as zero. Not sufficient.
Answer: E
08 Mar 2017, 10:26
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can we not cancel (x+1) from both the side?
Let's take 3 consecutive integers as x,(x+1),(x+2)so the product of them is x(x+1)(x+2)
(1) we have x+(x+1)+(x+2)=x(x+1)(x+2)
=>3x+3=x(x+1)(x+2)
=>3(x+1)=x(x+1)(x+2)
=>3=x(x+2)
x=-3 x=1
Let's take 3 consecutive integers as x,(x+1),(x+2)so the product of them is x(x+1)(x+2)
(1) we have x+(x+1)+(x+2)=x(x+1)(x+2)
=>3x+3=x(x+1)(x+2)
=>3(x+1)=x(x+1)(x+2)
=>3=x(x+2)
x=-3 x=1
