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

Note: $N$ is a single digit number.

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what ideas do you have on the problem? –  MSEoris Dec 24 '12 at 17:32
    
The tile should also change to use N. –  Amzoti Dec 24 '12 at 17:35
    
added a note. Thx –  harish.raj Dec 24 '12 at 17:39
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If N is a single digit number, and a positive integer, then there are only nine possibilities. You can exhaustively search the domain for the solution. A binary search could save you a step or two. –  Kaz Dec 24 '12 at 22:01
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5 Answers

up vote 4 down vote accepted

Start by trying to factorise the expression. One thing to note is that the sum of the coefficients is zero: $(+1) + (-7) + (+11) + (-5) = 0$, and so $n=1$ is a solution. We are able to factorise:

$$ n^3 - 7n^2 + 11n -5 \equiv (n-1)(\text{quadratic}) \, . $$

Assume the quadratic is of the form $a_2n^2+a_1n+a_0$, for some yet to be determined numbers $a_2$, $a_1$ and $a_0$. We may expand and compare coefficients. We see that $a_2 = 1,$ $a_1 - a_2 = -7$, $a_0-a_1 = 11$ and $-a_0 = -5.$ Solving these linear equations gives: $a_2 = 1,$ $a_1 = -6$ and $a_0 = 5$. Thus: $n^3-7n^2+11n + 5 \equiv (n-1)(n^2-6n+5).$ Again, we can factorise the quadratic term and we see that $n^2-6n+5 \equiv (n-1)(n-5)$. Thus:

$$n^3-7n^2+11n-5 \equiv (n-1)^2(n-5) \, . $$

The graph $y = x^3-7x^2+11x-5$ cuts the $x$-axis when $x=5$ and is tangent there when $x=1$. These are the critical values: $x=1$ and $x=5$. The question is: what happens when $x < 1$, when $1 < x < 5$ and when $x > 5.$ We can substitute values, say $x = 0$, $x = 2$ and $x = 6$.

$$ \begin{array}{ccc} 0^3 - 7\times 0^2 + 11\times 0 - 5 & = & -5 \, , \\ 2^3 - 7\times 2^2 + 11\times 2 - 5 & = & -2 \, , \\ 6^3 - 7\times 6^2 + 11\times 6 - 5 & = & 25 \, . \end{array} $$

It follows that $x^3-7x^2+11x-5 \le 0$ for all $x \le 5$ while $x > 0$ for all $x > 5$. Thus, for all integer $n > N = 5$ we have $n^3 - 7n^2 + 11n -5 > 0$.

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Just a typo, I think, but you say $n=0$ is a solution in the second line. I think it should be $n=1$ is a solution. –  user50407 Dec 24 '12 at 18:14
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The answer sought by this question is not $n = 6$. What the question asks is the greatest integer $N$, for which $n>N$ –  amWhy Dec 24 '12 at 18:16
    
@mr.FS Exactly right. Thanks for picking that up. I've corrected it. Thanks again. –  Fly by Night Dec 24 '12 at 23:17
    
@amWhy and FlyByNight: this is veering off-topic. Please consider moving to chat. –  robjohn Dec 25 '12 at 0:35
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$n^3-7n^2+11n-5=(n-1)(n^2-6n+5)=(n-1)^2(n-5)$

Now, for real $n,(n-1)^2 >0$ if $n \ne 1$

So, the sign of $(n-5)$ will determine the sign of the given polynomial if $n \ne 1$

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May I know the reason behind down voting? –  lab bhattacharjee Dec 24 '12 at 17:37
    
I have upvoted. I guess that the downvotes are for giving the complete answer, but thats not a good reason at all for downvoting a good answer. Downvoters note: the question isn't even tagged as homework! –  Belgi Dec 24 '12 at 17:38
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i think this version of the answer is much improved now and explains how the answer is found rather than just spits it out. nice edit. –  MSEoris Dec 24 '12 at 17:40
    
@Belgi, thanks for your feedback. But the down-vote persists. –  lab bhattacharjee Dec 24 '12 at 18:01
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$\text{We have}\quad f(n)=n^3-7n^2+11n-5\tag{1}.$

$\textrm{We want to solve for}\;\; n\;\textrm{ when}\;\;f(n) = n^3 - 7n^2 + 11n - 5 = 0\tag{2}$

Note that replacing $n$ with $1$ in equation $(2)$, and then summing, gives us: $$1(1^3) - 7(1^2 + 11(1) - 5 = 1 - 7 + 11 - 5 = 0,$$ and so $n = 1$ is a solution to $(2)$, and so $(n - 1)$ is at least one factor of the polynomial $(1)$.

Dividing $\;f(n) = n^3-7n^2+11n-5\;$ by $\;(n - 1)$ gives us $f(n)/(n-1) = (n^2 - 6n + 5).$

We are then able to factor $(1)$:

$$f(n)=(n-1)(n^2-6n+5)=(n-1)[(n-1)(n-5)]=(n-1)^2(n-5).\tag{3}$$

Now we look at the roots of $f(n)$; that is, when is it true that $f(n) = (n-1)^2(n-5) = 0\;$?. - We already know that when $n = 1$, $f(n) = 0$. - It is also the case that when $n = 5$, $f(n) = 0$.

So what must the value of $n$ be if we must have $f(n) > 0$?.

  • This is true when $n < 1$, but then, since $n$ is an integer, $n = 0$ is the greatest integer less than $1$, and we know that $n > N$, meaning $N \le -1$. But we need a positive $N$.
  • When $1 < n < 5$, $f(n) < 0.$ No good.
  • When $n > 5$, $f(n) > 0$. So we must have $n = 6$. This is the smallest positive integer for which $f(n)>0$.

Key Note: Now, knowing $n$ is NOT ENOUGH! This $n$ is NOT the answer to your question.

  • We need to use $n$ to determine the integer $N$. We want the smallest positive integer $N$ such that for any $n > N$, $f(n)$ is positive.
  • When $N = 5$, we have that for all $n > 5$, $f(n) > 0$: this is your answer to this question.
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Let $f(x)=x^3-7x^2+11x-5$. $f(1)=f(5)=0$ and so $$f(x)=(x-1)(x^2-6x+5)=(x-1)(x-1)(x-5)=(x-1)^2(x-5)$$ Then $$f(x)>0\iff ... $$

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The polynomial was chosen so that it would factor simply. So probably a solution through factoring is expected. But let's take a look at the polynomial $n^3-7n^2+11n-5$ as if factoring might be hard.

If $n\ge 7$, then $n^3-7n^2\ge 0$. And $11n-5\gt 0$ for any positive integer $n$. So $n^3-7n^2+11n-5\gt 0$ for any $n\ge 7$.

Now ask whether our polynomial is $\gt 0$ for $n\ge 6$. So plug in $n=6$. We get a positive result.

Is our polynomial $\gt 0$ for $n\ge 5$? Plug in $5$. We get $0$. This is not positive. So the required number $N$ is $5$.

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