This is in response to your question about the Day 16: Your Turn Explanation. I'm glad you asked, because the explanation is very interesting and a little abstract. When we discussed the last digit of a number, and a square, we were really talking about the remainder that we get after dividing that number by \(10.\) Now we are extending that idea to the case where instead of dividing by \(10,\) we are dividing by \(4.\) Let's look back to my comment in the Day 16: Challenge Discussions. In the same way that before we only cared about the last digit squared, and ignored the \(100 \times \text{ something} \) and \( 10 \times \text{ something } \) terms, here we only care about the remainder after dividing by \(4,\) so we can ignore the multiples of \(4,\) multiples of \(4^2,\) multiples of \(4^3,\) etc. We want to look at square numbers and the remainders that we get after dividing these squares by \(4.\) The clever trick is to look at just the remainders after dividing by \(4\): $$ (4 \times \text{ something } + R)^2 = 16 \times \text{ something } + 4 \times \text{ something } + R^2 $$ Something to be careful here is that the \(R \) isn't any single-digit number. It's only \(0, 1, 2 \text{ or } 3,\) because once we get \(R = 4,\) then it gets incorporated into the \(4 \times \text{ something } \) term instead. So since \(R\) can only be \(0, 1, 2, \text{ or } 3,\) that means \(R^2\) can only be \(0^2, 1^2, 2^2\) or \(3^2.\) These numbers are \(0, 1, 4\) or \(9.\) Since we care only about the remainder after dividing by \(4,\) let's remove \(4\) from the last two numbers. This means that \(R^2\) can be \(0, 1, 0,\) or \(1.\) Since we found that squares are equal to \(4 \times \text{ something } + R^2,\) and the first term is divisible by \(4,\) then the remainder of \(R^2\) (after dividing by \(4\)) is the remainder of the square number (after dividing by \(4\)). This means that squares either have remainders of \(0\) or \(1\) after dividing by \(4.\) The question asks us to add together two squares. The possibilities for the remainder of this sum (after dividing by 4) are: $$0 + 0 \\\\ 0 + 1 \\\\ 1 + 1 \\\\ $$ which give us \(0, 1 \) or \(2.\) It's impossible to get \(3.\) The question asks us for the impossible remainder, so the answer is \( \boxed{3.}\) I hope this helped! I'm really happy to answer any questions you have about the course content, no matter how big or small. I hope that you found this course beneficial to you and wish to learn more with us! Happy Learning, The Daily Challenge Team