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Each of the numbers a, b, c, d is equal to -1, 0, or 1. What is the [#permalink]
Bunuel wrote:
Bunuel wrote:

Tough and Tricky questions: Algebra.



We can see that \(a\) must equal 1; \(a\) is not 0 because there is no way for \(16b + 4c + d\) to add up to 44, and \(a\) is not -1 because there is no way for \(-64 +16b + 4c + d\) to add up to 44. By the same reasoning we used above, we know that \(d = 0\). Substitute to get: \(64 + 16b + 4c = 44\). Simplify: \(16b + 4c = -20\). Thus, \(b = -1\) and \(c = -1\).

We have solved for the value of each variable, so we can find the only possible value for the sum \(a + b + c + d\). Statement 2 is sufficient to answer the question.


Answer: B.


How did you conclude A=1 so quickly? Why not 2 or 3?

edit: haha oops, the answer to my question is given...

Originally posted by DangerPenguin on 13 Nov 2014, 09:58.
Last edited by DangerPenguin on 13 Nov 2014, 10:11, edited 1 time in total.
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Re: Each of the numbers a, b, c, d is equal to -1, 0, or 1. What is the [#permalink]
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DangerPenguin wrote:
Bunuel wrote:
Bunuel wrote:

Tough and Tricky questions: Algebra.



We can see that \(a\) must equal 1; \(a\) is not 0 because there is no way for \(16b + 4c + d\) to add up to 44, and \(a\) is not -1 because there is no way for \(-64 +16b + 4c + d\) to add up to 44. By the same reasoning we used above, we know that \(d = 0\). Substitute to get: \(64 + 16b + 4c = 44\). Simplify: \(16b + 4c = -20\). Thus, \(b = -1\) and \(c = -1\).

We have solved for the value of each variable, so we can find the only possible value for the sum \(a + b + c + d\). Statement 2 is sufficient to answer the question.


Answer: B.


How did you conclude A=1 so quickly? Why not 2 or 3?


We are given that each of the numbers a, b, c, d is equal to -1, 0, or 1.
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Re: Each of the numbers a, b, c, d is equal to -1, 0, or 1. What is the [#permalink]
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Re: Each of the numbers a, b, c, d is equal to -1, 0, or 1. What is the [#permalink]
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