# If $N = \prod \limits_ {k=1}^{40} \bigl(x-(2k-1) \bigr)^{2k-1}$ then for how many integer values of $x$ is $N$ negative?

If $N = \prod \limits_ {k=1}^{40} \bigl(x-(2k-1) \bigr)^{2k-1}$ then for how many integer values of $x$ is $N$ negative?

I am not getting any ideas for solving this one.

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When is a product of numbers negative? And if $a$ is positive/negative, what do you know about the positivity of an odd power $a^{2k-1}$ of $a$? – TMM Mar 3 '12 at 14:52

Hint: There is less to it than meets the eye. The powers are odd, so the sign of any term $(x-(2k-1))^{2k-1}$ is the same as the sign of $x-(2k-1)$. Our original polynomial, and the modified one where we replace each exponent by $1$, each have the roots $1$, $3$, $5$, and so on up to $79$. (You can keep the exponents if you like. It will make no real difference to the argument. But for me simplification is an automatic reflex.)
Look at a much smaller example, like $(x-1)(x-3)(x-5)(x-7)$. When $x<1$ or $x>7$, this is positive. The polynomial changes sign at each root. So we are negative at $2$, positive at $4$, negative at $6$, a total of $2$ integers at which we are negative. Now look at the same thing multiplied by $(x-9)(x-11)$. One more positive, one more negative, so there are $3$ integers for which our polynomial is negative. Now on to $40$ terms!
So, for $40$ terms the negatives will be at $x=2,6,10,14,\cdots 78$, That's $20$ terms Hence, $20$ integer solutions. – Quixotic Mar 9 '12 at 16:09
@Foool: Exactly. Here is a little elementary trick I use to count. We want to count how many numbers there are from $2$ to $78$, going up by $4$'s. This is the same as the number of numbers $4$ to $80$, going up by $4$'s. But that is the same as the number of numbers $1$ to $20$, going up by $1$'s (I just divided by $4$). This is clearly $20$. When we are counting, it is all too easy to be off by $1$, so one has to be cautious. Or else we can make a guess as to the formula, and check whether it works for a small number. – André Nicolas Mar 9 '12 at 16:19