Problem 7 · Ones and zeros divisible by 792
$792 = 8 \cdot 9 \cdot 11$: three divisibility rules at once.
Integer answer, at most 4 digitsHow many digits does the smallest number made of ones and zeros that is divisible by 792 have?
Reasoned solution
Key idea: $792 = 8 \cdot 9 \cdot 11$, and all three conditions translate into digit rules.
Divisible by $8$: the last three digits must form a multiple of $8$ made of ones and zeros: only $000$. Divisible by $9$: the digit sum (the number of ones) must be a multiple of $9$. Divisible by $11$: (ones in odd positions) $-$ (ones in even positions) $\equiv 0 \pmod{11}$; since that difference has the same parity as the total number of ones, $9$ ones is impossible ($9$ is odd and the difference would need to be $0$ or $\pm 11$). So $18$ ones are needed.
With $18$ consecutive ones and $000$ at the end everything holds ($9$ ones of each parity):