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If a and b are integers, is a^5 < 4^b ? (1) a^3 = –27 (2) b^2 = 16

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If a and b are integers, is a^5 < 4^b ? (1) a^3 = –27 (2) b^2 = 16  [#permalink]

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New post 26 Apr 2019, 02:38
10
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A
B
C
D
E

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  45% (medium)

Question Stats:

57% (01:11) correct 43% (01:13) wrong based on 405 sessions

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Re: If a and b are integers, is a^5 < 4^b ? (1) a^3 = –27 (2) b^2 = 16  [#permalink]

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New post 26 Apr 2019, 11:17
5
I thought statement 1 is sufficient. My answer is A.

Statement 2 is clearly insufficient.

Statement 1 establishes that a is negative.

No matter what integer you put as b in 4^b, it wouldnt return a negative value:
Ex:
4^4 = large positive number (256)
4^0 = 1 (still positive)
4^-2 = small fraction, but still positive (1/16)

Thus proving the inequality sound.

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Re: If a and b are integers, is a^5 < 4^b ? (1) a^3 = –27 (2) b^2 = 16  [#permalink]

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New post 26 Apr 2019, 10:33
From S1:

\(a^{-3} = 27\)
a = -3
But, No info about b.
Insufficient.

From S2:

\(b^2 = 16\)
b = +4 or -4
No info about a.
Insufficient.

Combining both:

-243 < 4^4
-243 < 256
Yes \(a^5 < 4^b\)
If b = -4
\(-243 < 4^{-4}\)
-243 < 1/256
Yes \(a^5 < 4^b\)

C is the answer.
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Re: If a and b are integers, is a^5 < 4^b ? (1) a^3 = –27 (2) b^2 = 16  [#permalink]

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New post 26 Apr 2019, 11:35
Chethan92 wrote:
From S1:

\(a^{-3} = 27\)
a = -3
But, No info about b.
Insufficient.

From S2:

\(b^2 = 16\)
b = +4 or -4
No info about a.
Insufficient.

Combining both:

-243 < 4^4
-243 < 256
Yes \(a^5 < 4^b\)
If b = -4
\(-243 < 4^{-4}\)
-243 < 1/256
Yes \(a^5 < 4^b\)

C is the answer.


Can you give me a scenario where 4^b will yield a negative value... I feel that A is sufficient because no matter what value b takes, 4^b will always be greater than 0 ...

Take extreme case , b = -1000/3 .. 4^-1000/3 = 4^-1/3(1000) = (1/4^1/3)^1000... I need to know what's a and statement 1 gives me value of a... Or atleast sign of a. Which is sufficient to tell that LHS<RHS... I would go with a... Am I missing something??

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Re: If a and b are integers, is a^5 < 4^b ? (1) a^3 = –27 (2) b^2 = 16  [#permalink]

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New post 26 Apr 2019, 11:37
1
Chethan92 wrote:
From S1:

\(a^{-3} = 27\)
a = -3
But, No info about b.
Insufficient.

From S2:

\(b^2 = 16\)
b = +4 or -4
No info about a.
Insufficient.

Combining both:

-243 < 4^4
-243 < 256
Yes \(a^5 < 4^b\)
If b = -4
\(-243 < 4^{-4}\)
-243 < 1/256
Yes \(a^5 < 4^b\)

C is the answer.


Can you provide me a scenario where 4^b will yield a negative value... I have been trying soo hard to figure out the RHS...

My reasoning -

A look at RHS - 4^b... If b = 0.. 4^0 = 1, if b = 1/2.. 4^1/2 = 2.. if b = -1/2 ... 4^-1/2 = 1/4^1/2 = 1/2... Take extreme case b= -1000/3 .. then 4^-1000/3 = 4^-1/3(1000)= {1/4^1/3}^1000....in every case, no Matter whether b is a negative integer, a positive integer, a fractional value... I get 4^b > 0... Soo I just need to know whether a is positive or negative... By statement 1 , a is negative... Which means LHS is <0 ... Which implies that LHS<RHS... sufficient...

In statement 2 - b^2 = 16... As explained earlier 4^b > 0... Nothing is mentioned about a... If a is negative integer, LHS < RHS... If a is positive integer LHS>RHS... I get two possibilities... Not sufficient... I would go with A... I don't know whether I missed out something.... I would appreciate if someone explain me what I did wrong??...

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Re: If a and b are integers, is a^5 < 4^b ? (1) a^3 = –27 (2) b^2 = 16  [#permalink]

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New post 26 Apr 2019, 11:38
Yes, correct. I missed it

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Re: If a and b are integers, is a^5 < 4^b ? (1) a^3 = –27 (2) b^2 = 16  [#permalink]

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New post 27 Apr 2019, 06:23
If base is negative & power is odd, resulting number must be negative.

And if base is positive, whatever the power, the resulting number must be POSITIVE.
Statement one is SUFFICIENT.

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Re: If a and b are integers, is a^5 < 4^b ? (1) a^3 = –27 (2) b^2 = 16  [#permalink]

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New post 27 Apr 2019, 06:32
Learning points:
If BASE is POSITIVE whatever the power (positive/negative/zero), the resulting number must be POSITIVE.

If base is Negative, there are two situations:
1. If power is ODD, the resulting number is NEGATIVE.
2. if power is EVEN, the resulting number is POSITIVE.


That means, there is a SINGLE situation in which the resulting number is NEGATIVE,
BASE IS NEGATIVE & POWER IS ODD.
In all other cases, the resulting number is POSITIVE.

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Re: If a and b are integers, is a^5 < 4^b ? (1) a^3 = –27 (2) b^2 = 16  [#permalink]

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New post 27 Apr 2019, 06:35
NEGATIVE POWER implies that the resulting number is FRACTION. (NEGATIVE POWER DOES NOT SAY ANYTHING WHETHER the resulting number is NEGATIVE or POSITIVE.)

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Re: If a and b are integers, is a^5 < 4^b ? (1) a^3 = –27 (2) b^2 = 16  [#permalink]

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New post 07 May 2019, 14:57
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Bunuel wrote:
If a and b are integers, is \(a^5 < 4^b\) ?

(1) \(a^3 = –27\)

(2) \(b^2 = 16\)


Target question: Is a^5 > 4^b

Statement 1: a³ = -27
Solve to get: a = -3
So, a^5 = (-3)^5 = -243
Since 4^b will be POSITIVE for all values of b, the answer to the target question is NO, a^5 is definitely NOT greater than 4^b
Since we can answer the target question with certainty, statement 1 is SUFFICIENT

Statement 2: b^2 = 16
Solve to get: EITHER b = 4 OR b = -4
Let's test some possible cases:
Case a: a = 1 and b = 4. In this case, a^5 = 1^5 = 1, and 4^b = 4^4 = 256. Here, the answer to the target question is NO, a^5 is definitely NOT greater than 4^b
Case b: a = 10 and b = 4. In this case, a^5 = 10^5 = 100,000, and 4^b = 4^4 = 256. Here, the answer to the target question is YES a^5 is greater than 4^b
Since we cannot answer the target question with certainty, statement 2 is NOT SUFFICIENT

Answer: A

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Re: If a and b are integers, is a^5 < 4^b ? (1) a^3 = –27 (2) b^2 = 16  [#permalink]

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New post 07 May 2019, 18:34
We need to determine whether a^5 is less than 4^b.

Statement One Alone:

a^3 = -27

We see that a is -3; so a^5 is negative. However, 4^b is always positive regardless of the value of b. Therefore, a^5 is indeed less than 4^b. Statement one alone is sufficient to answer the question.

Statement Two Alone:

b^2 = 16

So we see that b = 4 or -4.

However, without knowing anything about the value of a, we cannot determine whether a^5 is less than 4^b. For example, if a < 0, then a^5 < b^4. However, if a > 0, then a^5 might be greater than b^4 (for example, a = 5). Statement two alone is not sufficient to answer the question.


Answer: A
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Re: If a and b are integers, is a^5 < 4^b ? (1) a^3 = –27 (2) b^2 = 16  [#permalink]

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New post 10 May 2019, 16:26
Hi All,

We're told that A and B are INTEGERS. We're asked if A^5 is less than 4^B. This question can be approached with a mix of Number Properties and TESTing VALUES. To start, it's worth noting that raising +4 to ANY power will lead to a POSITIVE value (for example, 4^0 = 1, 4^-1 = 1/4, etc.).

(1) A^3 = -27

Fact 1 tells us that A = -3, although once you realize that A is a NEGATIVE value, you can stop working. By extension, A^5 would be a NEGATIVE value - and since 4^B is a POSITIVE value, we know that A^5 will ALWAYS be less than 4^B. Thus the answer to the question is ALWAYS YES.
Fact 1 is SUFFICIENT

(2) B^2 = 16

Fact 2 tells us that B = 4, but we don't know anything about the value of A. 4^4 = 16^2 = 256
IF...
A = 1, then A^5 = 1 and the answer to the question is YES.
A = 4, then 4^5 is clearly GREATER than 4^4, so the answer to the question is NO.
Fact 2 is INSUFFICIENT

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Re: If a and b are integers, is a^5 < 4^b ? (1) a^3 = –27 (2) b^2 = 16   [#permalink] 10 May 2019, 16:26
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