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

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Math Expert
Joined: 02 Sep 2009
Posts: 60626
If p is an integer and m= -p + (-2)^p, is m^3 >= -1  [#permalink]

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13 Nov 2019, 02:39
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Difficulty:

45% (medium)

Question Stats:

57% (02:55) correct 43% (02:33) wrong based on 37 sessions

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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

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Joined: 02 Aug 2009
Posts: 8335
If p is an integer and m= -p + (-2)^p, is m^3 >= -1  [#permalink]

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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) $$p^3 \leq -1$$

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 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, $$m= -(-2) + (-2)^{-2}=2+\frac{1}{4}$$

(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

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

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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, $$m= -(-2) + (-2)^{-2}=2+\frac{1}{4}$$

(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

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

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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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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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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

D

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