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!
