Question 880359
If the sum of the squares of the digits of a nine-digit positive integer is 2,
then that number has two digits that are 1, and seven digits that are 0.
To be a nine-digit positive integer, the first digit must be other than 0,
so the first digit must be 1.
Then the other digit that is also 1 could be the 2nd, 3rd, 4th, .... 8th or 9th digit.
That gives us {{{8}}} choices:
110,000,000,
101,000,000,
100,100,000,
100,010,000,
100,001,000,
100,000,100,
100,000,010, and
100,000,001.