# 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 ?

• Note that numbers, which are divisible by 11, also have an alternating digit sum that is divisible by 11, .i.e. $a-b+c\equiv 0\pmod{11}$. If you assume that the remainder must be $0$, you only have to consider the cases $a+c=b$ or $a+c=b+11$. – Reinhard Meier Jul 18 '17 at 11:47
• Technically, it doesn't say anything about the number being perfectly divisible by $11$. However, if the number isn't perfectly divisible by $11$, then when you divide by $11$, you cannot possibly get the sum of three prefect squares. So only the numbers divisible by $11$ are relevant regardless. – Arthur Jul 18 '17 at 12:00

## 4 Answers

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}

• It is a lot of work, !!! +1, thanks for spending time for the question – Atul Mishra Jul 22 '17 at 11:50

Make a table of all numbers, multiples of 11 from 110 to 990 as below:

There are two cases which satisfy the solution: 550 & 803. I did this simulation in excel.

n   a   b   c   a^2 b^2 c^2 sum n/11    f
550 5   5   0   25  25  0   50  50  1
803 8   0   3   64  0   9   73  73  1


EDIT: If I don't assume that all numbers would be multiples of 11, there are 8 solutions:

n   a   b   c   sum n/11 qoutient
131 1   3   1   11  11
241 2   4   1   21  21
324 3   2   4   29  29
550 5   5   0   50  50
624 6   2   4   56  56
803 8   0   3   73  73
900 9   0   0   81  81
910 9   1   0   82  82

• But I am doing this with pen and paper , computer program are not allowed @ProgSnob, but after your answer I too found 803 as a solution – Atul Mishra Jul 18 '17 at 12:14
• Ah, okay. Good luck! – ProgSnob Jul 18 '17 at 12:21

This might not be the best way but here is how I would do it.

Assume the remainder to be $\lambda$. Remember that $\lambda$ can take any value from 0 to 10.

Note that $100a + 10b +c = 11a^2 + 11b^2 + 11c^2 + \lambda$ can be rewritten as

$$(11a - 50)^2 + (11b - 5)^2 + (11c - \frac{1}{2})^2 = 2525 + \frac{1}{4} - \lambda$$

For the remainder of the answer, I'll assume $\lambda = 0.$ The other cases for $\lambda$ can be handled similarly. Since each of the terms on the left hand side are positive, we get that $c \leq 4$

We can take 5 cases based upon each of the five possible values for $c$ and solve for $a$ and $b$.

I won't go through all of the cases but in each one of them we should be able to form upper bounds on the values of $a$ and $b$. Furthermore, the search space can be pruned even further by making use of the parity of $a$ and $b$.

This method is not very different from your "table generating" method. The only difference is that the search space is pruned quite a bit. Also, if the number is itself divisible by 11, we can use the property mentioned in the comments to prune the search space even more.

The process can be repeated for other values of $\lambda$. The upper bounds should not differ a lot but the individual solutions will.

• Yes @iamwhoiam , but I still cannot find the numbers which leaves some remainder when divided by 11 through this method , can you suggest me how can I proceed for them?? – Atul Mishra Jul 18 '17 at 14:48
• @AtulMishra, I've updated my answer. I'm quite sure that there are smarter ways to prune the search space than what I've suggested – iamwhoiam Jul 18 '17 at 15:02

The following method can be implemented without using a computer, which is (as far as I understand) what you are looking for.

As Reinhard Meier points out, if

\begin{align*} 100a + 10b + c = 11(a^2 + b^2 + c^2), \end{align*}

which is an integer, then by considering the left-hand side (mod $11$), one finds that \begin{align*} 100a + 10b + c \equiv a - b + c \equiv 0, \end{align*} meaning that $a + c \equiv b$ (mod $11)$. Since $-9 \leqslant a-b+c \leqslant 18$, the only possibilities are the following: Either $a-b+c =0$ or $a-b+c = 11$. We thus have two cases:

\begin{align*} 1) \quad a+c = b, \quad \text{or} \quad 2) \quad a+c = 11+b. \end{align*}

Case 1) -- Writing $a+c$ for $b$ in the equation, one arrives at the following statement: $110a + 11c = 22a^2 + 22c^2 + 22ac$, which reduces to the following quadratic equation in $a$: $a^2 + a(c-5)+c^2-\frac{c}{2}=0.$ The discriminant of the equation is

\begin{align*} D = (c-5)^2 -4(c^2-\frac{c}{2}) = 25-3c^2-8c, \end{align*}

and this has to be a square number, since we require a solution $a$ to belong to the set $\lbrace 0,\ldots,9 \rbrace$. One can now check easily that $D$ is a square iff $c = 0$. In that case, $a + 0 = a = b$, and so the initial equation becomes $110a = 22a^2$, where the only solutions are $a= 0$ or $a = 5$. This yields the number $550$, which you mention in your post.

Case 2) -- Writing $a+c -11$ for $b$, we have \begin{align*} b^2 &= (a+c)^2 -22(a+c)+121 \\ &= a^2 + 2ac + c^2 -22a-22c + 121 \\ &= a^2 + a(2c-22) + c^2 - 22c + 121. \end{align*}

The original equation now says

\begin{align*} 110a+11c-110 &= 11a^2 + 11c^2 + 11(a^2 + a(2c-22) + c^2 - 22c + 121)\\ &\Updownarrow \\ 10a+c-10 &= a^2 + c^2 + a^2 + a(2c-22) + c^2 - 22c + 121 \\ &= 2a^2 +a(2c-22)+2c^2-22c+121\\ &\Updownarrow \\ 0 &= a^2 + a(c-16)+c^2-\frac{23}{2}c+\frac{131}{2}. \end{align*} Just like before, the discriminant of the quadratic equation needs to be a square number. The discriminant is

\begin{align*} D &= (c-16)^2 - 4(c^2-\frac{23}{2}c+\frac{131}{2}) \\ &= c^2 -32c + 256 -4c^2+46c-262 \\ &= 14c - 3c^2 - 6. \end{align*}

One easily verifies that for $c \geqslant 5$, $D < 0$. Hence we only need to consider the possibilities $c = 0,1,2,3,4$. Out of these, only $c = 3$ yields a square discriminant, namely $D = 9$. Returning to the original equation, this now says that

\begin{align*} 0 = a^2 -13a + 9-\frac{69}{2}+\frac{131}{2} = a^2 -13a +40, \end{align*} whose solutions are $a = 5$ (and thus $b = 5+3-11 < 0$, which is impossible) or $a = 8$ (and thus $b = 8+3-11 = 0$).

All in all, we find two (and only two) numbers that satisfy the specified requirement: 550 and 803.

• Yes @BaconAndX , but you have quite misread the question, I know there are two numbers550 and 803 which are divisible by 11 and satisfying property, but I need help in finding those numbers which are not divisible by 11 , but still satisfy the property,as mentioned by ProgSnob in his answer, there are 6 such numbers, how can I find them without using computer program – Atul Mishra Jul 21 '17 at 1:06