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# m is the smallest positive integer such that

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Retired Moderator
Joined: 25 Feb 2013
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m is the smallest positive integer such that  [#permalink]

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12 Sep 2017, 09:46
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95% (hard)

Question Stats:

38% (02:43) correct 62% (02:16) wrong based on 85 sessions

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$$m$$ is the smallest positive integer such that for any integer $$n≥m$$, the quantity $$n^3–7n^2+11n–5$$ is positive. What is the value of $$m$$?

A. 4
B. 5
C. 8
D. 6
E. 11
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Joined: 15 Mar 2017
Posts: 38
Location: India
GMAT 1: 720 Q50 V37
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Re: m is the smallest positive integer such that  [#permalink]

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12 Sep 2017, 10:59
niks18 wrote:
$$m$$ is the smallest positive integer such that for any integer $$n≥m$$, the quantity $$n^3–7n^2+11n–5$$ is positive. What is the value of $$m$$?

A. 4
B. 5
C. 8
D. 6
E. 11

on factorizing the equation, we get $$(n-1)^2$$(n-5)>0

For this equation to always be positive the range of n should be from (-∞,1)U(5,∞)
The smallest positive integer that will satisfy the value will be 6
Hence n=m=6
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You give kudos, you get kudos. :D
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Joined: 13 Jan 2017
Posts: 34
Re: m is the smallest positive integer such that  [#permalink]

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12 Sep 2017, 14:03
D.

Let f(n)=n^3–7n^2+11n–5 then

f(4)<0 - need to calculate
f(5)=0 - need to calculate
f(6)>0 - need to calculate, correct answer

f(8)>0 as (n^3–7n^2)>0 and (11n–5)>0 - obvious, no need to calculate
f(11)>0 similarly to f(8) case
Retired Moderator
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Re: m is the smallest positive integer such that  [#permalink]

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12 Sep 2017, 21:46
SaGa wrote:
niks18 wrote:
$$m$$ is the smallest positive integer such that for any integer $$n≥m$$, the quantity $$n^3–7n^2+11n–5$$ is positive. What is the value of $$m$$?

A. 4
B. 5
C. 8
D. 6
E. 11

on factorizing the equation, we get $$(n-1)^2$$(n-5)>0

For this equation to always be positive the range of n should be from (-∞,1)U(5,∞)
The smallest positive integer that will satisfy the value will be 6
Hence n=m=6

Hi SaGa

the highlighted portion is not correct. the equation will not be positive for any values of $$n<5$$ and at $$n=1,5$$, the equation will be $$0$$
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Joined: 02 Aug 2009
Posts: 7958
Re: m is the smallest positive integer such that  [#permalink]

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12 Sep 2017, 22:07
1
1
niks18 wrote:
$$m$$ is the smallest positive integer such that for any integer $$n≥m$$, the quantity $$n^3–7n^2+11n–5$$ is positive. What is the value of $$m$$?

A. 4
B. 5
C. 8
D. 6
E. 11

Hi..

Firstly in gmat, the choices are in ascending or descending order....

Now for solution.
In these Q, substitution is the best way as equation is complicated.
And in substitution start from the centre digit which would help you eliminate two choices below or above it

Now $$n^3-7n^2+11n-5=n^2(n-7)+11n-5$$
Now the above situation will be positive for n as 7 or above for sure, so check for values below it.
Also as mentioned above in red colour, check for middle value in the choices.

For n as 6, $$6^2(6-7)+11*6-5=-36+61>0$$.... possible value
n as 5, $$5^2(5-7)+11*5-5=25*(-2)+55_5=0$$.. so when n=m=5, ans is 0... NO..

Ans is 6..
D
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m is the smallest positive integer such that  [#permalink]

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13 Sep 2017, 11:39
niks18 wrote:
$$m$$ is the smallest positive integer such that for any integer $$n≥m$$, the quantity $$n^3–7n^2+11n–5$$ is positive. What is the value of $$m$$?

A. 4
B. 5
C. 8
D. 6
E. 11

OE

Given: $$n^3–7n^2+11n–5$$$$>0$$, or

$$n^3-6n^2-n^2+5n+6n-5>0$$, or

$$n^3-6n^2+5n-n^2+6n-5>0$$

$$n(n^2-6n+5)-1(n^2-6n+5)>0$$. this implies $$(n-1)(n^2-6n+5)>0$$, this can be further factorized and we finally get

$$(n-1)^2(n-5)>0$$---------(1).

Now $$(n-1)^2$$ will be always positive, hence for equation (1) to be positive $$n-5>0$$

or $$n>5$$. Hence the smallest integer, satisfying the inequality is $$6$$. so $$n≥6$$ and as $$n≥m$$. So $$m=6$$

Option D

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I see from previous discussions that people are finding it hard to re-arrange the equation and factorize so presenting a solution -

One of the method that is very handy in factorizing a bit complex equation is to get one of the roots of the equation, as it will provide one of the factor of the expression. you can use the factor to divide the expression and get other factors. For e.g in this question

$$f(n)=n^3-7n^2+11n-5$$, the root of the function will be a value where $$f(n)=0$$. In GMAT, expression will be fairly simple, hence one root can be identified very easily. if you just ignore the $$"n"$$ in the expression and add it, you will realize that the summation is $$0$$ i.e. one of the root of the expression $$1$$.

so, $$f(1)=0$$, this means $$n=1$$ is one root of the equation.

so $$n-1=0$$ or $$(n-1)$$ will be the factor of the equation. divide the equation by $$(n-1)$$, you will get quotient as $$n^2-6n+5$$.

so our equation can be written as $$(n-1)(n^2-6n+5)$$. Now it becomes fairly simple to factorize

Not a big fan of solving through options as GMAT 700 level questions can hardly be solved through options and moreover it takes away the fun out of Mathematics
m is the smallest positive integer such that   [#permalink] 13 Sep 2017, 11:39
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