Determine all three-digit numbers N having the property that N is divisible by 11, and N/11 is equal to the sum of the squares of the digits of N.

Question

Determine all three-digit numbers N having the property that N is divisible by 11, and N/11 is equal to the sum of the squares of the digits of N.

Answer 1

To start, it's important to define some variables for the 3-digit number set let's define the set A = {100,101,102...,997,998,999}.

Make A the domain of a function f(x), such that f(x) = x mod 11, and now for all value of x which f(x)=0, we have the set S defined as {10,11,12,13...87,88,89,90}

(realize that neither 9 nor 91 is not contained in the set because and 9 x 11 =99 and 91 x 11 = 1001, are not 3-digit numbers)

Now, in order to define which are the square digits of n let's define our 3-digit number in function of our new set S.

S = {10,11,12...87,88,89,90} where S is set set of numbers in the format "xy", where x and y are natural numbers

eg.S = {xy} = {10} => x=1,y=0 and so on.

11 S = {110,121,132,...,968,979,990}

Multiply the "literal" set by 11 xy x 11

and separate between the decimal units

if (x+y)<10

C	D	U
x	(x+y)	y

or

C	D	U
x+1	(x+y)mod 10	y

and the sum the square of those digits must be equal to xy itself

and we'll have 2 cases

#1st case for this case, we'll have to choose 2 digits 0-9, such that there's no zero on the first digit and the sum of those digits is less than 10 , which gives us a total of 45 comparisons. but now let's find a way to make those faster

C	D	U
x	(x+y)	y

x²+(x+y)²+y² = "xy" (the quotation marks are to represent that xy is a number not a product)

Expanding... x²+x²+2xy+y²+y²= "xy" => 2(x²+xy+y²)= "xy" =>

x²+xy+y² = $\frac{"xy"}{2}$

as we defined earlier x and y are natural numbers and this means that y must be divisible by 2 which reduces ours comparisons to 25.

we can also make this number of comparisons smaller once we start iterating setting a value to x, starting at 1

for x=1

1²+y+y²= $\frac{"1y"}{2}$ (where "1y" could be {10,12,14,16,18})

but before the comparison you could test if the number is in the desired range by doing y(y+1) $\le$ $\frac{(x+1)"0"}{2}$ -x²

y(y+1) $\le$ $\frac{(1+1)"0"}{2}$-1 =>

y(y+1) $\le$ 10-1 => y(y+1) $\le$ 9, and with this affirmation we can reduce the number of possibilities for 1, from 5 to only 2 which are:

for y=0

y(y+1)= $\frac{"1y"}{2}$-1 => 0(0+1)= $\frac{"10"}{2}$-1 => 0 = 4 false

for y=2

y(y+1)= $\frac{"1y"}{2}$-1 => 2(2+1)= $\frac{"12"}{2}$-1 => 6 = 5 false

for x=2

y(y+1) $\le$ $\frac{(2+1)"0"}{2}$ -2² and with this affirmation we can reduce the number of possibilities for 2, from 4 to 3(from now on this approach is not effective anymore to use) which are:

for y=0

y(y+1)= $\frac{"2y"}{2}$-4 => 0(0+1)= $\frac{"20"}{2}$-4 => 0 = 6 false

for y=2

y(y+1)= $\frac{"2y"}{2}$-4 => 2(2+1)= $\frac{"22"}{2}$-4 => 6 = 7 false

for y=4

y(y+1)= $\frac{"2y"}{2}$-4 => 4(4+1)= $\frac{"24"}{2}$-4 => 20 = 8 false

for x=3

for y=0

y(y+1)= $\frac{"3y"}{2}$-9 => 0(0+1)= $\frac{"30"}{2}$-9 => 0 = 6 false

for y=2

y(y+1)= $\frac{"3y"}{2}$-9 => 2(2+1)= $\frac{"32"}{2}$-9 => 6 = 7 false

for y=4

y(y+1)= $\frac{"3y"}{2}$-9 => 4(4+1)= $\frac{"34"}{2}$-9 => 20 = 8 false

for x=4

for y=0

y(y+1)= $\frac{"4y"}{2}$-16 => 0(0+1)= $\frac{"40"}{2}$-16 => 0 = 4 false

for y=2

y(y+1)= $\frac{"4y"}{2}$-16 => 2(2+1)= $\frac{"42"}{2}$-16 => 6 = 5 false

for y=4

y(y+1)= $\frac{"4y"}{2}$-16 => 4(4+1)= $\frac{"44"}{2}$-16 => 20 = 4 false

for x=5

for y=0

y(y+1)= $\frac{"5y"}{2}$-25 => 0(0+1)= $\frac{"50"}{2}$-25 => 0 = 0 TRUE

for y=2

y(y+1)= $\frac{"5y"}{2}$-25 => 2(2+1)= $\frac{"52"}{2}$-25 => 6 = 1 false

for y=4

y(y+1)= $\frac{"5y"}{2}$-25 => 4(4+1)= $\frac{"54"}{2}$-25 => 20 = 2 false

for x=6

for y=0

y(y+1)= $\frac{"6y"}{2}$-36 => 0(0+1)= $\frac{"60"}{2}$-36 => 0 = -6 false

for y=2

y(y+1)= $\frac{"6y"}{2}$-36 => 2(2+1)= $\frac{"62"}{2}$-36 => 6 = -5 false

Since we're receiving negative values as output now, let's assume that no other value greater than 6 will reach our goal so, let's move on to the final step

as we have tested for values smaller than 6 let's take a guess and start the second case at 6

#2nd case (x+y)$\ge$ 10

we will have the following equation:

(x+1)² + [(x+y) mod 10]²+ y² = "xy"

for x=6 [(6+y) mod 10]²+ y² = "6y" - (6+1)²

for y=4

[(6+y) mod 10]²+ y² = "6y" - 49
[(6+4) mod 10]²+ 4² = "64" - 49 => 16 = 15 false

for y=5

[(6+y) mod 10]²+ y² = "6y" - 49
[(6+5) mod 10]²+ 5² = "65" - 49 => 26 = 16 false

for y=6

[(6+y) mod 10]²+ y² = "6y" - 49
[(6+6) mod 10]²+ 6² = "66" - 49 => 40 = 17 false

as we've also seen by this sequence the left side is growing must faster than the right side, therefore, it will never be equal, let's move on for the next value of x

for x=7 [(7+y) mod 10]²+ y² = "7y" - (7+1)²

for y = 3

[(7+y) mod 10]²+ y² = "7y" - 64
[(7+3) mod 10]²+ 3² = "73" - 64² => 9 = 9 TRUE

for y = 4

[(7+y) mod 10]²+ y² = "7y" - 64
[(7+4) mod 10]²+ 4² = "74" - 64² => 17 = 10 false

for the same reason than the last one the left side now is growing faster than the right side so they will never meet again

for x=8

[(8+y) mod 10]²+ y² = "8y" - (8+1)²

for y = 2

[(8+y) mod 10]²+ y² = "8y" - 9²
[(8+2) mod 10]²+ 2² = "82" - 81 => 4 = 1 false

from now on the left side side of the equation will always be greater than the right one so, we have found every number that multiplied by eleven, the sum of their digits squared equals itself.

The number we found out were = {50,73}

To set them in the format of the question, all you gotta do is multiply them by 11

and there you go

50x11 = 550 73x11 = 803

Solution = {550, 803}

I hope this answer is useful to you, the resolution got more extensive than I imagined, and the great challenge of questions like this is to answer them fast, you can learn new methods to go faster, but the most important part is training. Thank you for the question. Please, if there's any new question or suggestion in order to make my resolutions better please let me know.

Answer 2

I think there are only two - 550 and 803. and maybe 0.

Answer 3

I think I might have a simpler solution than the one above. Suppose $N = 100 \cdot a + 10\cdot b + c$ and $0 \leqslant a,b,c \leqslant 9 $. We know two things: $$ I) \hspace{1cm} 100a + 10b +c \equiv 0 \mod 11 $$ $$ II) \hspace{1cm} N = 11a^2 + 11b^2 +11c^2$$ From $I)$, we get that $99a + a +11b-b+c \equiv a - b +c \equiv 0 \mod 11$, which implies that $b = a+c$ or $b = a+c-11$.

From $II)$, we get that $100 \cdot a + 10\cdot b + c = 11a^2 + 11b^2 +11c^2$.

Then you can test for the equation above in the two cases $b = a+c$ or $b = a+c -11$.

For the first case in particular you'll see that you can use that one side o the equation being even implies that the other side is also even. This reduces the total number of tests to $15$, I think $(0,2,4,6,8$ for the first case and $0$ through $9$ for the second case).