What number is n if n-1 is divisible by 15 and n+1 divides 1001?

Question

Does anyone know how I could solve this problem in my number theory course. I am kind of clueless about it. Got some ideas but none of them gives some concrete answer. I figured out that n is probably 76 but that is just by guessing the answer.

Answer 1

$1001=7\cdot 11\cdot 13$, so the divisors of $1001$ are $$n+1\in\{1, 7, 11, 13, \underbrace{77}_{7\cdot 11}, \underbrace{91}_{7\cdot 13}, \underbrace{143}_{11\cdot 13}, 1001\}$$ Removing $2$ to all divisors we have
$$n-1\in\{-1, 5, 9, 11, 75, 89, 141, 999\}$$ Dividing by $15$ we have $$\frac{n-1}{15}\in\left\{\frac{-1}{15}, \frac{5}{15}, \frac{9}{15}, \frac{11}{15}, 5, \frac{89}{15}, \frac{141}{15}, \frac{999}{15}\right\}$$ The only integer is $5$, so $$\dfrac{n-1}{15}=5\;\Rightarrow\; n=76$$

Answer 2

We find in the comments that $76$ is correct. We can identify it without trial and error as follows:

Since $(n+1)|(1001=7×11×13)$, we must have

$n+1=7^a×11^b×13^c$

where $a,b,c\in\{0,1\}$. Evidently $n+1\equiv2\bmod15$ and so must end in $2$ or $7$ base ten. Terminal digit $2$ is impossible for an odd number and terminal digit $7$ requires $a-c=1$ ($3\equiv7^{-1}\bmod10$). So $a=1,c=0$. To get $b$ we add the condition $n+1\equiv2\bmod3$, implied by the $\bmod15$ residue, forcing $b=1$.

Thereby $n+1=77$ and the answer follows.

Answer 3

$n-1=15N\Rightarrow n-1\equiv 0\space\text { or } 5\pmod{10}\Rightarrow n\equiv 1\space\text { or } 6\pmod{10}$.

$n+1$ is not even because it is a factor of $1001$. Then $n$ should be even so $n=10k+6$ and $n+1=10k+7$. Since the divisors of $1001$ are $1,7,11,13,77,91,143,1001$ the only possibilities are $n+1=7$ and $n+1=77$.

Thus $\color{red}{n=76}$