Find all positive integers $n$ such that $72|(10^n+8)$

Question

We can see that $n\ge3$ by objection.

How to prove this?

The sum of digits is $9$, hence the number is divisible by $9$. The last $3$ digits are $008$, hence the number is divisible y $8$. — barak manos, Nov 20 '16 at 13:55

score 2 · Accepted Answer · answered Nov 20 '16 at 13:57

2

Sum of digits of $10^n+8$ is $9$, hence $9|10^n+8$.

$n\geq3\implies10^n+8\equiv8\pmod{1000}\implies8|10^n+8$.

Therefore $n\geq3\implies(9|10^n+8)\wedge(8|10^n+8)$.

$\gcd(9,8)=1\implies9\cdot8|10^n+8\implies72|10^n+8$.

answered Nov 20 '16 at 13:57

barak manos

43,109

score 1 · Answer 2 · answered Nov 20 '16 at 14:09

First, $10^n + 8 \equiv 1^n + 8 \equiv 0$ mod(9) implies that $9 \mid 10^n + 8$ for any positive integer n. Next, $10^n + 8 \equiv 2^n \equiv 0$ mod(8) is valid if n is greater than or equal to 3. So, we have that 72 divides this quantity for n greater than or equal to 3.

Find all positive integers $n$ such that $72|(10^n+8)$

2 Answers2