Same Digit in All Place Values

Let $n$ have $k$ digits, all equal to $d$ .

For $8$ to divide $n$ , we need $8$ to divide the last three digits of $n$ , that is $\overline{ddd}=d \cdot 111$ . The only possibility is $d=8$ .

For $9$ to divide $n$ , we need $9$ to divide the sum of digits of $n$ , that is $8k$ ; since $\gcd(8,9)=1$ , this means $9$ divides $k$ .

For $7$ to divide $n$ , we need $6$ to divide $k$ . This is because $7$ divides $1001$ , so $7$ divides $111111$ .

Note that these three divisors are enough; for instance, if both $8$ and $9$ divide $n$ , then $6$ does too, and so on.

The smallest $k$ that works is $18$ ; so the smallest $n$ is a string of eighteen $8$ s, that is $n=\boxed{888888888888888888}$ .

As every digit is same of the number, and the number is divisible by 8, we can say the number is 88888... upto some n digits.. So I use brute force to find the n or the length of the number..

1
2
3
4

for i in range(1,20):
        n = int("8"*i)
        if n%7 == 0 and n%9 == 0:
            print(i)

1

18

[Hello, this is not a complete solution :) ]

Related Topics: converting repeating decimal into a fraction and multiplicative order.

To generalize the solution

Let N be a same-digit-repeating-integer, and $k$ is the least common multiple of all given divisors of N. The prime factorization of $k$ is $2^{a_0}.3^{a_1}.5^{0}.7^{a_3}...p_z^{a_z}, (a_0<4)$ . Let $2^{a_0}= s$

If $s= 4 or 8$ then $\displaystyle N = s \times \frac{10^{lcm( O( 10, \frac{k}{s}), 3^{a_1})} -1 }{9}$ .

$O(a,b)$ returns the multiplicative order of $a \mod b$

So here we have $k= lcm(1, 2, 3, 4, 6, 7, 8, 9)= 2^3 \times 3^2 \times 7= 504; s=2^3=8; O(10, \frac{504}{8})= 6$

So $N=8 \times \frac{ 10^{18} -1}{9}= 888888888888888888$