How many int. values of n will the expression be greater than 1 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 ?
 A: 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 $48$.
A: 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.
A: 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.
A: $\rm 2 \le 4n\!+\!7 \le 199\! \iff\! 3 \le 4(n\!+\!2) \le 200.\:$ 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 $\lfloor 200/4\rfloor = 50$.
