Find the sum of all integer values of the expression $\frac{7n+12}{2n+3}$ for integer values of $n$

Question

The expression $\dfrac{7n+12}{2n+3}$ takes integer values for certain integer values of $n$. What is the sum of all such integer values of the expression?

My attempt: $$\frac{7n+12}{2n+3} = 3 + \frac{n+3}{2n+3}$$ then finding all integers $k$ that $$k=\frac{n+3}{2n+3}$$ and trying out all suitable values; is there other ways solving without enumeration?

Hint: If $k$ is an integer, so is $2k-1$. That leaves very few possibilities for $n$, and you can brute force the rest. But A) where is the question from? Did you search the site? — Jyrki Lahtonen, Feb 02 '25 at 16:35
Welcome to Mathematics Stack Exchange. Note that $2(7n+12)-7(2n+3)=3$ — J. W. Tanner, Feb 02 '25 at 16:35
there is no positive integers less than 1, @ChristopheBoilley — zenin1, Feb 02 '25 at 16:35
@ChristopheBoilley It was not specified that the integers should be positive! — Jyrki Lahtonen, Feb 02 '25 at 16:37
@zenin1 Then it will be difficult to find an integer $k$ such that $k=\frac{n+3}{2n+3}$. — Christophe Boilley, Feb 02 '25 at 16:39
Questions similar to this appear in contests, like this classic. So you should specify the origin. — Jyrki Lahtonen, Feb 02 '25 at 16:39
This isn't the best kind of a duplicate, because there only common factors (rather than integer values) were searched after. There's more... this, but still not constrained to simplify to an integer... — Jyrki Lahtonen, Feb 02 '25 at 16:43
@JyrkiLahtonen thank you, actually I searched question but find only hight math problems, It seems I didn't search correctly — zenin1, Feb 02 '25 at 16:47

score 3 · Accepted Answer · answered Feb 02 '25 at 16:44

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Since $2(7n+12)-7(2n+3)=3$, if $2n+3$ divides $7n+12$, then $2n+3$ divides $3$.

That means $2n+3\in\{\pm1,\pm3\}$. Can you take it from here?

answered Feb 02 '25 at 16:44

J. W. Tanner

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how it becomes $2(7n+12)−7(2n+3)=3$ – zenin1 Feb 02 '25 at 16:58
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$2(7n+12)-7(2n+3)=14n+24-14n-21=3$ – J. W. Tanner Feb 02 '25 at 17:02
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Please strive not to post more (dupe) answers to dupes of FAQs (& PSQs), cf. site policy announcement here. – Bill Dubuque Feb 02 '25 at 21:05
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In the purported dupe, the leading coefficient of one of the polynomials was $1$, so it didn’t need to be multiplied by anything, unlike the situation here – J. W. Tanner Feb 03 '25 at 00:13

score 0 · Answer 2 · answered Feb 02 '25 at 18:35

If you have already found that $n = -3, -2, -1, 0$ which gives $k = 0, -1, 2, 1$ respectively, you can show that there must be no further integer values of $n$ by considering the graph. Note that:

$$f(n) := \frac{n + 3}{2n + 3} = \frac{1}{2} + \frac{3/2}{2n + 3}$$

so the horizontal asymptote is $n = \frac{1}{2}$. As this is a suitable transformation of the graph of $g(n) = \frac{1}{n}$, $f(n)$ must also be decreasing (with an asymptote where $2n + 3 = 0 \Rightarrow n = -\frac{3}{2})$.

Given the values of $k$ above, we have shown that $k$ is not an integer for $n \in (-\infty, -3) \cup (0, \infty)$, as there is no integer in $(0, \frac{1}{2})$ nor in $(\frac{1}{2}, 1)$. Hence proven.

Please strive not to post more (dupe) answers to dupes of FAQs (& PSQs), cf. site policy announcement here. — Bill Dubuque, Feb 02 '25 at 21:05

score -1 · Answer 3 · answered Feb 02 '25 at 19:46

It should be obvious that if $n > 0$ then $\frac {n+3}{2n+3} < 1$ and can't be an integer. (As $2n+3,n+3$ are both positive we'd have $0 < \frac {n+3}{2n+3} < 1$).

So for $n \le 0$.... Well, it's pretty obvious if $n =0$ we have $k =\frac {n+3}{2n+3} =1$.

And if $n< 0$ .....

Well.... $\frac {n+3}{2n+3} = \pm \frac {|n+3|}{|2n+3|}$ and for that to be an integer we need $|n+3| \ge |2n+3|$ which should strike us as very unlikely as coeficient of $2$ will dominate for large values of $|n|$.

That is to say. If $n < 0$ and $|n|$ is large enough so that both $n+3$ and $2n+3$ are both negative (that is if $n < -3$ or $|n| > 3$) we would have

$\frac {n+3}{2n+3} =\frac {|n+3|}{|2n+3|}=\frac{-n-3}{-2n-3}= \frac {|n|-3}{2|n| - 3}$. But as $0< |n|-3< 2|n|-3$ and both terms are positive that means $0 <\frac{|n|-3}{2|n|-3} < 1$ and it can't be an integer.

So $n > 0$ is impossible and $n<-3$ is impossible.

So we have to test for $n =-3,-2,-1,0$. Which....are easy to see yield

$k = \frac {-3+3}{2\cdot(-3)+3}=0; k = \frac {-2+3}{2(-2)+3}=-1; k=\frac {-1+3}{-2+3}=2; k=\frac {0+3}{2\cdot 0 + 3} = 1$.

Please strive not to post more (dupe) answers to dupes of FAQs (& PSQs), cf. site policy announcement here. — Bill Dubuque, Feb 02 '25 at 21:05

Find the sum of all integer values of the expression $\frac{7n+12}{2n+3}$ for integer values of $n$

3 Answers3