Find all natural numbers $n$ such that $n+1$ divides $3n+11$

Question

Following the example of my teacher:

Find all natural numbers $n$ such that $n-2$ divides $n+5$.

$$n+5 = n-2+7$$

As $n-2|n-2$, $n-2$ will divide $n+5$ if and only if $n-2|7$. Yet, $7$ has the divisors $-7$, $-1$, $1$, $7$ hence the equations:

$n-2=-7 \Leftrightarrow n = -5 $

$n-2=-1 \Leftrightarrow n = 1 $

$n-2=7 \Leftrightarrow n = 3 $

$n-2=-1 \Leftrightarrow n = 9 $

So $S = \{ 1, 3, 9 \}$

I decompose $n+1$ the exact same way:

$$n+1 = 3n+11 - 2(n+5)$$

But I'd get stuck as $2(n+5)$ since only $-2$, $-1$, $1$ and $2$ are divisors, which don't satisify the equation as I'd hoped:

$3n+11 = -2 \Leftrightarrow n =\frac{-13}{3}$
$3n+11 = -1 \Leftrightarrow n = -4$
$3n+11 = 1 \Leftrightarrow n =\frac{-10}{3}$
$3n+11 = 1 \Leftrightarrow n = -3$

Any clues?

What has $n+5$ to do with this? – Angina Seng Sep 21 '19 at 06:59 — Angina Seng, Sep 21 '19 at 06:59
It's more intuitive if you use congruences - see my answer. – Bill Dubuque Sep 21 '19 at 13:31 — Bill Dubuque, Sep 21 '19 at 13:31

score 3 · Accepted Answer · answered Sep 21 '19 at 07:00

3

$n+1$ certainly divides $3n+3$. If it divides $3n+11$, it must also divide $(3n+11)-(3n+3)$.

answered Sep 21 '19 at 07:00

Angina Seng

158,341

score 2 · Answer 2 · answered Sep 21 '19 at 07:05

2

You can use polynomial long division, which is useful if you have a more complex expression:

$$ \require{enclose} \begin{array}{rll} 3 && \\[-3pt] n+1 \enclose{longdiv}{\ 3n+11}\kern-.2ex \\[-3pt] \ \ - \underline{\ (3n+3)} && \\[-3pt] 8 && \\[-3pt] \end{array} $$

Therefore $\frac{3n+11}{n+1} = 3 + \frac{8}{n+1}$. Can you continue?

answered Sep 21 '19 at 07:05

Toby Mak

16,827

1

+1 for being a much clearer explanation than the teachers and for being the first time I've seen long division in LaTeX. – Sep 21 '19 at 10:01
Thanks man! I personally learnt it this way, but I prefer synthetic division which is probably out of reach for the OP here. The MathJax was copied from this post. – Toby Mak Sep 21 '19 at 10:12
I just taught synthetic division to my AP Calc students this week. It isn't so bad as long as you can look at it alongside polynomial long division and understand that the same numbers show up in both diagrams and why the subtraction got to change to addition. – Sep 21 '19 at 10:51

score 1 · Answer 3 · answered Sep 21 '19 at 07:05

1

Set $n+1=m$

$3n+11=3(m-1)+11=3m+8\equiv8\pmod m$

answered Sep 21 '19 at 07:05

lab bhattacharjee

274,582

score 1 · Answer 4 · answered Sep 21 '19 at 07:24

When the example has $$ n + 5 = n-2 + 7 $$ it might have been more clearly written $$ n + 5 = 1\cdot(n-2) + 7 $$ to highlight the quotient, $1$, and remainder, $7$, of the division $\frac{n+5}{n-2}$.

For your problem, $$ 3n + 11 = 3 \cdot (n+1) + 8 \text{.} $$ That is, the quotient is $3$ and the remainder is $8$. So you want to inspect the choices of $n$ such that $(n+1) \mid 8$.

score 1 · Answer 5 · answered Sep 21 '19 at 13:30

1

$\!\! \bmod n\!+\!1\!:\,\ \color{#c00}{n\equiv -1}\,\Rightarrow\, 3\,\color{#c00}n+11\equiv 3(\color{#c00}{-1})+11\equiv 8\ $ by Congruence Sum & Product Rules $\ \ \ \ \ $

answered Sep 21 '19 at 13:30

Bill Dubuque

272,048

Find all natural numbers $n$ such that $n+1$ divides $3n+11$

5 Answers5