Multiple of 17 sums to 17

A number and its digit sum are always congruent mod $9$ ; since $17$ and $9$ are relatively prime, it follows that $N \equiv 1 \pmod{9}$ .

The largest 4-digit number whose digit sum is $17$ is $9800$ , so $N \le \left\lfloor \dfrac{9800}{17} \right\rfloor = 576$ . Since $N \equiv 1 \pmod{9}$ , it follows that $N \in \{568, 559, 550, ....\}$ . We try

• $568 \times 17 = 9656 \$ but $\ 9 + 6 + 5 + 6 = 26$

• $559 \times 17 = 9503 \$ and $\ 9 + 5 + 0 + 3 = 17$

so the required $N$ is $559$ and our answer is $559 + 9503 = \boxed{10062}$

Inspired problem

---

Divisibility rule of 17

Subtract 5 times the last of digit from the remaining truncated number. If the result is divisible by 17 then the original number is also divisible by 17.

---

Call $n =\overline{abcd}$ the largest four digit number. As per the divisibility test of $17$ . We have new number $m =\overline{abc} -5d =10^2a+10b +c -5d$ . Assume number is divisible by $17$ . Then we get $17 m =10^2a+10b+c-5d = 99a + 9b -6d +\overbrace{(a+b+c+d)}^{\text{ 17}} \\17(m-1)=3(33a+3b-2d)\implies 17 \left(\dfrac{m-1}{3} \right) = 33a+3b-2d$ Further simplifying we see $b = \dfrac{1}{3}\left[17\left(\dfrac{m-1}{3}\right) -33a +2d\right] < 30, m> 4$ For the largest value of $\overline{abcd}$ we want to maximize the value of $a =9,$ . Also we notice that $0\leq b\leq 8$ and $\small{281 \leq 17\left(\dfrac{m-1}{3}\right)\leq 305\implies 50 \leq m\leq 54}$ or $\small{297 \leq 17\left(\dfrac{m-1}{3}\right)\leq 321 \implies 53 \leq m \leq 57}$ Hence, we have $50\leq m\leq 57$ , the largest possible for $m$ is $55$ such that $b\in\mathbb N \leq 8$ .

Plugging back to equation above $\small{b = \dfrac{306-297+2d}{3}= \dfrac{9+2d}{3}\implies d= 0 , 3 , 6}$ which solves for $b = 5, 5 ,7$ and largest 4 digits number we have are $\overline {abcd} = 9530, 9503, 97c6$ (rejected).

Therefore, the largest possible value of $\overline{abcd} = 9503$ which is equal to $17\times 559$ . Final result of $N+\overline{abcd} =\boxed{ 10062}$ .