Hmmmm

nope...misread Q

Sorry I'm not CU, but I'll try this question. I've done a variant of it before I think.

Let's call N = 100a + 10b + c

Then we know:

100a + 10b + c = 11m for some arbitrary integer m

m = a^2 + b^2 + c^2 

100a + 10b + c = 11a^2 + 11b^2 + 11c^2.

Using divisibility rule of 11, we know:

Case 1:

a-b+c = 0

a+c = b

or Case 2: 

a-b+c = 11

b = a+c-11

this splits the problem into two cases for us:

1. b = a +c where the first case for divisibility of 11 holds true.

100a + c + 10a + 10c = 11a^2 + 11c^2 + 11(a+c)^2

10a + c = 2a^2 + 2ac+ 2c^2

10a + c = 2a^2 + 2ac + 2c^2

Here, we're kind of stuck. Realize that we can factor out 2 from the right hand side, meaning that c must be even since a is also even.

we can write c = 2z, with z being either 0, 1, 2, 3, or 4, and substitute it in

we get:

10 a + 2z = 2a^2 + 4az + 8q^2

Dividing by 2 on both sides, this gives us:

5a + z = a^2 + 2az + 4z^2

We just bash out cases now, and eventually we find:

when we have z = 0, the case works out, with a being 5, and c = 0  b = 5. From this case, we get that N = 550. All other cases don't work(2,4,6,8)

I'll leave the rest of the problem to you; there's only one other case!

2. ?????

also edit: sorry for the bad "typography". Had to do this in a hurry

Also just wanted to put this out there. If you have lots of time to do this problem, I'd actually recommend just bashing out all of the possibilities divisible by 11. Because there are only ~ 80 - 81(do the math?) divisors of 11 from 100 - 999, I would recommend listing all of them out and then testing on a case by case basis(of course, this heavily relies on you having lots of time). 

Let N=110a+10b+c for some digits a,b, and c . Thenfor some .

We also have m=a^2+b^2+c^2. 

For an integer divisible by 11, the sum of digits in the odd positions minus the sum of digits in the even positions is divisible by 11.

Substituting, we get 100a+10b+c=11a^2+11b^2+11c^2

Case 1: b=1+c.

100a+c+10a+10c=11a^2+11c^2+11(a+c)^2.

10a+c=2a^2+2ac+2c^2

Case 2: Substitute q= 0-4 and solve for a

q=0: This works.

q=1: This is not an integer. This doesn't work for a.

q=2: This is not an integer. This doesn't work for a.

q=3: This is not an integer. This doesn't work for a.

q=4: This is not an integer. This doesn't work for a.

Another solution: There are 900 three-digit numbers. Only 81 of these numbers are divisible by 11. Now, list the answer. We end up getting N=550,803.

Hope this helped!

Welp I'm late I just woke up 3 hours ago and its 2:30 at my location. Nice work guys

Thank you!!