Problem 7 · 45a with all equal digits
A repdigit multiple of 45 must end in 5 and have digit sum divisible by 9.
Integer answer, at most 4 digitsThe number $a$ is a positive integer and the sum $S = a + 2a + 3a + 4a + \dots + 9a$ is a number whose digits are all equal. What is the smallest possible value of $S$? As your answer give a two-digit number $\overline{bc}$, where $b$ is how many times the repeated digit appears and $c$ is that digit. For example, if $S = 77777$, the answer is 57, since there are five sevens.
Reasoned solution
Key idea: $S = (1+2+\cdots+9)\,a = 45a$: it is a multiple of $45$, i.e. of both $5$ and $9$.
A number with all digits equal to $d$: to be a multiple of $5$ it must end in $5$ (if $d = 0$ it isn't positive), so $d = 5$. To be a multiple of $9$, the digit sum $5k$ must be a multiple of $9$: since $5$ and $9$ are coprime, $k$ must be a multiple of $9$.
The smallest is $k = 9$:
Answer in $\overline{bc}$ format: $b = 9$ fives, $c = 5$ → $95$.