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How would I solve this problem:

How many integer values of n will the expression $4n+7$ be an integer greater than 1 and less than 200 a)48 b)49 c)50 d)51 e)52 ? Ans 50

I am trying to solve this by using inequalities and doing something like this

$200 > 4n + 7 > 1$

$193 > 4n > -6 $

$\frac{193}{4} > n > \frac{-6}{4}$

However this just gives me the range. How do i get the number of integers 50 ?

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You need to count the number of integers that satisfy the inequalities. – copper.hat Jul 30 '12 at 18:28
Well, if integer then $\,n\,$ can be negative as well... – DonAntonio Jul 30 '12 at 18:30
Your right hand side should be $-\frac{6}{4}$. – copper.hat Jul 30 '12 at 18:31
Take some time, make a list: $3,7,11,\color{green}{.}\color{goldenrod}{.}\color{red}{.},199$ and count! – draks ... Jul 30 '12 at 18:39
I think this one has been beaten to death... – copper.hat Jul 30 '12 at 19:32
up vote 6 down vote accepted

We can work with inequalities as you did. We want $$1\lt 4n+7 \lt 200,$$ which can be rewritten as $$-6 \lt 4n \lt 193,$$ and then as $$\frac{-6}{4} \lt n \lt \frac{193}{4}.$$ Note that $\,\frac{-6}{4}=-1.5\,$ and $\,\frac{193}{4}=48.25\,$ So all integers from $-1$ to $48$ inclusive work, and no others. There are $\,50 \,$ of them: the integers $-1$ and $0$, plus the $48$ integers from $1$ to $48$.

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Thanks for the great tip ! – MistyD Jul 30 '12 at 19:19
The middle equation should be $4n$, not $n$. – SiliconCelery Jul 30 '12 at 19:20
@SiliconCelery: Thanks, fixed. I added that middle line late, and used paste/correct, with insufficient correct. – André Nicolas Jul 30 '12 at 19:24

n is (as you said) an integer. So for $\frac{193}{4} > n > \frac{-6}{4}$, it follows $48 > n > -1 $.

So for $n \in \{-1, ... , 48\}$ the inequalties hold true. Those are exactly 50 integers.

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In general, there are exactly $b-a$ integers between $a$ and $b$, including $b$ but not $a$. For example, there are exactly $10-0=10$ integers between 0 and 10, including 10 but not 0. If you include both ends, add 1 to the total; if you include neither, subtract 1.

You have:

$$-\frac64 < n < \frac{193}4$$

which is:

$$-1\frac12 < n < 48\frac14$$

So $n$ can be any integer between $-1$ and $48$, inclusive; there are $48 - (-1) + 1 = 50$ such.

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$\rm 2 \le 4n\!+\!7 \le 199\! \iff\! 3 \le 4(n\!+\!2) \le 200.\:$ The map $\rm\:n\to 4(n\!+\!2)\:$ bijects all solutions $\rm\:n\:$ with all multiples of $4$ that are $\ge 3$ and $\le 200$, of which there are $\lfloor 200/4\rfloor = 50$.

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