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Is c^(abcd) = 1? (1) |a| > |b| > |c| > |d| (2) c^d > 3

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Is c^(abcd) = 1? (1) |a| > |b| > |c| > |d| (2) c^d > 3  [#permalink]

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New post 02 Dec 2019, 01:19
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A
B
C
D
E

Difficulty:

  55% (hard)

Question Stats:

44% (01:42) correct 56% (01:31) wrong based on 43 sessions

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Re: Is c^(abcd) = 1? (1) |a| > |b| > |c| > |d| (2) c^d > 3  [#permalink]

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New post 02 Dec 2019, 02:45
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Bunuel wrote:
Is \(c^{abcd}=1\)?

(1) \(|a|>|b|>|c|>|d|\)

(2) \(c^d>3\)


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We need to find if \(c^{abcd}=1\) or not
Let’s do some pre-analysis before analyzing each statement.
    • If c = 1, then \(c^{abcd}\) will be 1.
    • If c = -1 and {abcd} is even, then also we’ll get \(c^{abcd} = 1\)
    • If abcd = 0 and c ≠ 0, then even in that case \(c^{abcd} = 1\)
Keeping the above in mind, let’s look at the first statement.

Statement 1: \(|a|>|b|>|c|>|d|\)
    • From this statement, we can infer that magnitude wise, a, b, c, and d are all distinct.
      o And since all modulus value is positive or 0, we can be sure that c ≠ 0, since |c| > |d|
Now, let’s take two scenarios:
Case 1: d =0
    • Then \(c^{abcd} = c^0 = 1\)
Case 2: d ≠ 0
    • Then \(c^{abcd}\) may or may not be equal to 1.
Thus, statement 1 is not sufficient to answer the question.

Statement 2: \(c^d>3\)
• This statement, clearly tell us that c ≠ 1 or -1
• Also, d ≠ 0
However, we don’t know any about a and b, if any one turns out to be 0, then \(c^{abcd}\) could turn out to be 1.
Thus, statement 2 is also not sufficient to answer the question.

Combining both statements:
• |a| > |b| > |c| > |d|
• d ≠ 0 and c ≠ 1 or -1
If d ≠ 0, then that means a, b, or c are also not equal to 0. Thus, abcd ≠ 0
Also, since c ≠ 1 or – 1, this helps us infer that \(c^{abcd} ≠ 1\).
Thus, the correct answer is Option C.
GMAT Club Bot
Re: Is c^(abcd) = 1? (1) |a| > |b| > |c| > |d| (2) c^d > 3   [#permalink] 02 Dec 2019, 02:45
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