STEP II 2020 Digit sum

First, the digit sum of a $k$ -digit number is at most $9k$ , whereas a $k$ -digit number is (by definition) at least $10^{k-1}$ . We therefore need $44\cdot 9k = 396k \ge 10^{k-1}$

It's easy to check that this means we are interested in numbers with at most $4$ digits.

So, let's put $n=1000a+100b+10c+d$ (where $a,b,c,d$ are all in the range $0$ to $9$ inclusive). The digit sum of $n$ is $a+b+c+d$ ; so $\begin{aligned} 44a+44b+44c+44d &=1000a+100b+10c+d \\ 956a+56b &=34c+43d \end{aligned}$

The right-hand side is at most $34\cdot 9+43\cdot 9=693$ which forces $a=0$ and $56b=34c+43d$

Taking the equation modulo $9$ , we find $b+c+d \equiv 0$ ; so $n$ is a multiple of $9$ . Also, $n$ is a multiple of $44$ ; hence $n$ is a multiple of $396$ and is less than $1000$ . There are only two options; $n=396$ doesn't work, but $n=792$ does; so $n=\boxed{792}$ is the unique solution.

If $n$ is a positive integer such that $n = 44DS(n)$ , where $DS(n)$ is the sum of the digits of $n$ , then certainly $n$ is a multiple of $44$ . If $n$ has $X$ digits then $10^{X-1} \; \le \; n \; = \; 44DS(n) \; \le \; 44 \times 9X \; = \; 396X$ so that $10^X \; \le \; 3960X$ which implies that $X \le 4$ . There are no $1$ -digit numbers with this property. Since $44$ and $88$ are the only $2$ -digit multiples of $44$ , there are no $2$ -digit numbers with this property.

Now consider a $3$ -digit number $n = \overline{abc}$ . Then $100a + 10b + c = 44(a + b + c)$ , so that $56a = 34b + 43c$ . This means that $c = 2\gamma$ is even, and that $28a = 17b + 43\gamma$ . Since $n$ is a multiple of $11$ , $a - b + c \equiv 0 \pmod{11}$ , and so $b = a + c - 11e$ where $e = 0$ or $1$ . Thus $28a = 17a + 17c - 187e + 43\gamma = 17a + 77\gamma - 187e$ , so that $a = 7\gamma - 17e$ . Since $7\gamma = a + 17e \; \le \;a + 17 \le 26$ we deduce that $\gamma \le 3$ . There are four possibilities:

• $\gamma=0$ gives $e=0$ and $a=0$ , which is impossible,
• $\gamma = 1$ gives $e=0$ and $a=7$ , $b=9$ and $c=2$ ,
• $\gamma=2$ gives $a=14 - 17e$ , which is impossible,
• $\gamma=3$ gives $e=1$ and $a=4$ , $b=-1$ and $c=6$ , which is impossible

The only $3$ -digit number with the desired property is $792 = 44(7 + 9 + 2)$ .

Now consider a $4$ -digit number $n = \overline{abcd}$ . Then $n = 1000a + 100b + 10c + d = 44(a + b + c +d)$ , so that $956a + 56b \; = \; 34c + 43d$ In addition $a-b+c-d \equiv 0 \pmod{11}$ , so that $b + d = a + c + 11e$ where $e = -1$ , $0$ or $1$ . But then $n = 44(a + b + c + d) =88(a + c) + 484e \; \le \; 88 \times 18 + 484 = 2068$ , and hence $a \le 2$ . But then $n \le 88 \times 11 + 484 = 1452$ , so it follows that $a = 1$ . But this means that $956 \; \le \; 956a + 56b \; = \; 34c+ 43d \; \le \; (34 + 43)9 = 693$ which is impossible. Thus there are no $4$ -digit numbers with the required property.

Thus the only positive integer with the desired property, and hence the sum of all integers with that property, is $\boxed{792}$ .