number theory - Solutions to $100a+10b+c=11a^2+11b^2+11c^2$ and $100a+10b+c=11a^2+11b^2+11c^2+ k$

The problem states that

Find all three digit natural numbers such that when the number is divided by $11$ gives the quotient equal to sum of squares of their digits

Since there is no information about whether remainder is $0$ or not , I firstly assumed that the question is talking about the numbers perfectly divisible by $11$

Now I have $80$ numbers left , I can check them separately , but it will be lengthy

I made an equation $$100a+10b+c=11a^2+11b^2+11c^2$$

Rearranged and got $$a(11a-100)+b(11b-10)+c(11c-1)=0$$

I have $10$ values for $a,b,c=\{1,2,3,4,5,6,7,8,9,0\}$

I made a table corresponding to $a(11a-100),b(11b-10),c(11c-1)$ and found that only for $a=b=5,c=0$ and $a=8,b=0,c=3$ are giving their sum $0$ , hence $550$ and $803$ are the only numbers satisfying given property and divisible by 11.

Now I have two questions:

$1.)$Is my way and my answer correct? If no, then where have I misunderstood?

$2.)$What about the numbers which are not divisible by $11$?

As mentioned by an answer er of this post , there are six such numbers which are not divisible by $11$ , but still give the quotient the sum of square of their digits. But answer er found it using a computer program, which is not suitable for pen paper mathematics. So how can I find those six numbers ?

Answer

We want to find all sets of integers $a,b,c,k$ such that $$100a+10b+c=11a^2+11b^2+11c^2+ k\tag1$$
where $1\le a\le 9,0\le b\le 9,0\le c\le 9$ and $0\le k\le 10$.

We have
$$(1)\iff k+b(11b-10)+c(11c-1)=a(100-11a)$$

Since $a(100-11a)\le 5(100-11\times 5)=225$, we have
$$k+b(11b-10)+c(11c-1)\le 225\tag2$$

Since we have that $b(11b-10)\ge 6(11\times 6-10)=336\gt 225$ for $b\ge 6$ and that $c(11c-1)\ge 5(11\times 5-1)=270\gt 225$ for $c\ge 5$, from $(2)$, we have to have $b\le 5$ and $c\le 4$.

Also, from $(1)$ with $0\le k\le 10$, solving $$0\le 100a+10b+c-11a^2-11b^2-11c^2\le 10$$for $a$ gives$$a\in\left[\frac{50-\sqrt m}{11},\frac{50-\sqrt{m-110}}{11}\right]\cup\left[\frac{50+\sqrt{m-110}}{11},\frac{50+\sqrt m}{11}\right]\tag3$$where $m=2500-11(11b^2+11c^2-10b-c)$.

Now, we see that $$\frac{50-\sqrt m}{11}\le 1\le \frac{50-\sqrt{m-110}}{11}\iff 1521\le m\le 1631\quad\text{for}\quad a=1$$$$\frac{50-\sqrt m}{11}\le 2\le \frac{50-\sqrt{m-110}}{11}\iff 784\le m\le 894\quad \text{for}\quad a=2$$$$\frac{50-\sqrt m}{11}\le 3\le \frac{50-\sqrt{m-110}}{11}\iff 289\le m\le 399\quad\text{for}\quad a=3$$$$\frac{50-\sqrt m}{11}\le 4\le\frac{50-\sqrt{m-110}}{11}\iff 110\le m\le 146\quad\text{for}\quad a=4$$$$\frac{50+\sqrt{m-110}}{11}\le 5\le\frac{50+\sqrt m}{11}\iff 100\le m\le 125\quad \text{for}\quad a=5$$$$\frac{50+\sqrt{m-110}}{11}\le 6\le\frac{50+\sqrt m}{11}\iff 256\le m\le 356\quad\text{for}\quad a=6$$$$\frac{50+\sqrt{m-110}}{11}\le 7\le\frac{50+\sqrt m}{11}\iff 729\le m\le 829\quad\text{for}\quad a=7$$$$\frac{50+\sqrt{m-110}}{11}\le 8\le \frac{50+\sqrt m}{11}\iff 1444\le m\le 1544\quad\text{for}\quad a=8$$$$\frac{50+\sqrt{m-110}}{11}\le 9\le\frac{50+\sqrt m}{11}\iff 2401\le m\le 2501\quad\text{for}\quad a=9$$

Case 1 : When $b=5$, since $b(11b-10)=225$ with $(2)$, we have to have $k=c(11c-1)=0,a=5$ giving $a=5,b=5,c=0,k=0$.

Case 2 : When $b=4$, since $b(11b-10)=136$,

$$(2)\implies k+c(11c-1)\le 225-136=89$$from which we have to have $c\le 2$ since $c(11c-1)\ge 3\times (11\times 3-1)=96\gt 89$ for $c\ge 3$.

• Case 2-1 : For $c=0$, we have $m=1004$. So, there is no integer $a$ satisfying $(3)$.




• Case 2-2 : For $c=1$, we have $m=894$. So, we have $a=2$, and $k=10$ from $(1)$.




• Case 2-3 : For $c=2$, we have $m=542$. So, there is no integer $a$ satisfying $(3)$.

Case 3 : When $b=3$, since $b(11b-10)=69$,
$$(2)\implies k+c(11c-1)\le 225-69=156$$from which we have to have $c\le 3$ since $c(11c-1)\ge 4\times (11\times 4-1)=172\gt 156$ for $c\ge 4$.

• Case 3-1 : For $c=0$, we have $m=1741$. So, there is no integer $a$ satisfying $(3)$.




• Case 3-2 : For $c=1$, we have $m=1631$. So, we have $a=1$, and $k=10$ from $(1)$.




• Case 3-3 : For $c=2$, we have $m=1279$. So, there is no integer $a$ satisfying $(3)$.




• Case 3-4 : For $c=3$, we have $m=685$. So, there is no integer $a$ satisfying $(3)$.

Case 4 : When $b=2$, we have $c\le 4$.

• Case 4-1 : For $c=0$, we have $m=2236$. So, there is no integer $a$ satisfying $(3)$.




• Case 4-2 : For $c=1$, we have $m=2126$. So, there is no integer $a$ satisfying $(3)$.




• Case 4-3 : For $c=2$, we have $m=1774$. So, there is no integer $a$ satisfying $(3)$.




• Case 4-4 : For $c=3$, we have $m=1180$. So, there is no integer $a$ satisfying $(3)$.




• Case 4-5 : For $c=4$, we have $m=344$. So, we have $a=3,6$, and $(a,b,c,k)=(3,2,4,5),(6,2,4,8)$ from $(1)$.

Case 5 : When $b=1$, we have $c\le 4$.

• Case 5-1 : For $c=0$, we have $m=2489$. So, we have $a=9$ and $k=8$ from $(1)$.




• Case 5-2 : For $c=1$, we have $m=2379$. So, there is no integer $a$ satisfying $(3)$.




• Case 5-3 : For $c=2$, we have $m=2027$. So, there is no integer $a$ satisfying $(3)$.




• Case 5-4 : For $c=3$, we have $m=1433$. So, there is no integer $a$ satisfying $(3)$.




• Case 5-5 : For $c=4$, we have $m=597$. So, there is no integer $a$ satisfying $(3)$.

Case 6 : When $b=0$, we have $c\le 4$.

• Case 6-1 : For $c=0$, we have $m=2500$. So, we have $a=9$, and $k=9$ from $(1)$.




• Case 6-2 : For $c=1$, we have $m=2390$. So, there is no integer $a$ satisfying $(3)$.




• Case 6-3 : For $c=2$, we have $m=2038$. So, there is no integer $a$ satisfying $(3)$.




• Case 6-4 : For $c=3$, we have $m=1444$. So, we have $a=8$, and $k=0$ from $(1)$.




• Case 6-5 : For $c=4$, we have $m=608$. So, there is no integer $a$ satisfying $(3)$.

Therefore, the answer is
$$\begin{align}\color{red}{(a,b,c,k)=}\ &\color{red}{(5,5,0,0),(2,4,1,10),(1,3,1,10),(3,2,4,5)}\\&\color{red}{(6,2,4,8),(9,1,0,8),(9,0,0,9),(8,0,3,0)}\end{align}$$