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The integers A, B, and C are consecutive and A < B < C. If A^2 = C

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V
Joined: 02 Sep 2009
Posts: 50772
The integers A, B, and C are consecutive and A < B < C. If A^2 = C  [#permalink]

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New post 12 Oct 2017, 23:05
00:00
A
B
C
D
E

Difficulty:

  15% (low)

Question Stats:

78% (00:51) correct 22% (00:54) wrong based on 95 sessions

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The integers A, B, and C are consecutive and A < B < C. If A^2 = C, which of the following could be the value of A?

I. –1
II. 0
III. 2

A. I only
B. III only
C. I and II only
D. I and III only
E. I, II, and III

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Re: The integers A, B, and C are consecutive and A < B < C. If A^2 = C  [#permalink]

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New post 13 Oct 2017, 04:06
Bunuel wrote:
The integers A, B, and C are consecutive and A < B < C. If A^2 = C, which of the following could be the value of A?

I. –1
II. 0
III. 2

A. I only
B. III only
C. I and II only
D. I and III only
E. I, II, and III


\(A+2 = A^2\)
\(A^2 - A - 2 = 0\)
\((A-2)(A+1) = 0\)
\(A = 2\) \(or -1\)

Answer is D.
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Re: The integers A, B, and C are consecutive and A < B < C. If A^2 = C  [#permalink]

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New post 16 Oct 2017, 16:08
Bunuel wrote:
The integers A, B, and C are consecutive and A < B < C. If A^2 = C, which of the following could be the value of A?

I. –1
II. 0
III. 2

A. I only
B. III only
C. I and II only
D. I and III only
E. I, II, and III


Let’s test out Roman numerals to see if we have A^2 = C.

I. If A = -1, then B = 0 and C = 1. Since (-1)^2 = 1, Roman numeral I works.

II. If A = 0, then B = 1 and C = 2. Since 0^2 = 0 ≠ 2, Roman numeral II does not work.

III. If A = 2, then B = 3 and C = 4. Since 2^2 = 4, Roman numeral III works.

Answer: D
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Re: The integers A, B, and C are consecutive and A < B < C. If A^2 = C &nbs [#permalink] 16 Oct 2017, 16:08
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The integers A, B, and C are consecutive and A < B < C. If A^2 = C

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