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If p is an integer and m= -p + (-2)^p, is m^3 >= -1

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If p is an integer and m= -p + (-2)^p, is m^3 >= -1  [#permalink]

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New post 13 Nov 2019, 02:39
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If p is an integer and m= -p + (-2)^p, is m^3 >= -1  [#permalink]

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New post 13 Nov 2019, 05:48
Bunuel wrote:
If p is an integer and \(m= -p + (-2)^p\), is \(m^3 \geq -1\)

(1) p is even

(2) \(p^3 \leq -1\)


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\(m= -p + (-2)^p\)
If p is even, m will always be positive...
If p<0 and even, -p and (-2)^p both will be positive, so m is positive
If p<0 and odd, -p is positive and (-2)^(-p) will be negative but a fraction that will be less than -p



(1) p is even
Let p=0, \(m= -p + (-2)^p=-0+(-2)^0=0+1=0\)
Let p=-2, \(m= -(-2) + (-2)^{-2}=2+\frac{1}{4}\)
m>0...answer is YES

(2) \(p^3 \leq -1\)
means \(p \leq -1\)..
\(m= -p + (-2)^p\)... both -p and \((-2)^p\)are positive if p is even.
\(m= -p + (-2)^p\)=positive+positive=positive
If p is odd -p will be positive and greater than (-2)^-p

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Re: If p is an integer and m= -p + (-2)^p, is m^3 >= -1  [#permalink]

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New post 13 Nov 2019, 15:30
Hello Sir!

I have a doubt for statement "If p<0, -p and (-2)^p both will be positive, so m is positive"
Take p = -3
(-2)^-3= -1/8 which is negative.

But the sum of -p+(-2)^p will always be positive as we move towards more negative numbers.


chetan2u wrote:
Bunuel wrote:
If p is an integer and \(m= -p + (-2)^p\), is \(m^3 \geq -1\)

(1) p is even

(2) \(p^3 \leq -1\)


Are You Up For the Challenge: 700 Level Questions



\(m= -p + (-2)^p\)
If p is even, m will always be positive...
If p<0, -p and (-2)^p both will be positive, so m is positive



(1) p is even
Let p=0, \(m= -p + (-2)^p=-0+(-2)^0=0+1=0\)
Let p=-2, \(m= -(-2) + (-2)^{-2}=2+\frac{1}{4}\)
m>0...answer is YES

(2) \(p^3 \leq -1\)
means \(p \leq -1\)..
\(m= -p + (-2)^p\)... both -p and \((-2)^p\)..
\(m= -p + (-2)^p\)=positive+positive=positive

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Re: If p is an integer and m= -p + (-2)^p, is m^3 >= -1  [#permalink]

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New post 13 Nov 2019, 16:15
Let p=-2, m=−(−2)+(−2)−2=2+1/4

I'm a bit confused with this line. Shouldn't it be 2 - 1/4?
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Re: If p is an integer and m= -p + (-2)^p, is m^3 >= -1  [#permalink]

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New post 13 Nov 2019, 16:23
Square or even power of any number or digit is always non-negative.


minustark wrote:
Let p=-2, m=−(−2)+(−2)−2=2+1/4

I'm a bit confused with this line. Shouldn't it be 2 - 1/4?
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If p is an integer and m= -p + (-2)^p, is m^3 >= -1  [#permalink]

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New post 15 Nov 2019, 00:01
worked on stem a bit,

m^3 >= -1
means -1<=m < infinity
or is -1<= -p+ (-2)^p < infinity ??????


A substitue -1/2 -5 and 3 , all satisfy hence correct.

B p^3 <= -1 means p<= -1 . Hence -p = positive + (-2)^ something negative = positive + 1/(-2)^positive .

Also -p will always be greater than 1/ (-2)^p
If p is even negative lets say -2 expression becomes positive + 1/(-2) positive = positive + positive .
Hence yes m >= -1

If p is odd negative lets say -1 expression becomes 1 -1 = 0 .
Hence yes m>= -1

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If p is an integer and m= -p + (-2)^p, is m^3 >= -1   [#permalink] 15 Nov 2019, 00:01
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