Somehow this led us to divisibility tests. She knew the one for 2 (and I assume she knows the similar tests for 5 and 10). Most people know these tests. My dad was making nachos (for some reason we were doing her homework on the kitchen floor) and overheard me teaching her the divisibility test for 4. He got all indignant: "I didn't know that! Why don't they teach us cool things like that in school?" and ran into the other room, from where we heard his shouts to Kathi, "Did you know that?!?"
The divisibility test for 4 has the same justification as the test for 2 (or, for that matter, for 5 or 10). For 2, you can write your (positive, for the sake of argument) lengthy integer as 10X + Y where Y is less than 10 (and greater than or equal to 0). Then, if Y is divisible by 2 (so Y = 2Y') we have
so the big integer is also divisible by 2. (Note that this works because of the distributive property mentioned earlier.) And, likewise, if Y is not even then the big long integer is also not even. For 4, you instead write your integer as 100X + Y, where Y is now less than 100. Since 4 divides 100, you only have to check whether 4 divides Y (so now you have to check whether 4 divides a 2 digit number, which is more work than the even/odd test, but considerably less work than trying to divide 4 into, say, a 7 digit long number).
So Emma tested some big long numbers for divisibility by 4 and after we found a few that were not divisible by 4, I had her come up with a big long number that was divisible by 4. She ended her number with 08, which I liked.
Then I taught her the divisibility test for 3. If you add up the digits of the number and that sum is divisible by 3, then the whole number is divisible by 3. Rather than give a proof, which is much cleaner and nicer if you know about modular arithmetic, I will just give an example. This also uses the distributive property Emma was working on.
635,148 = 6(100,000) + 3(10,000) + 5(1,000) + 1(100) + 4(10) + 8
=6(1+99,999) + 3(1+9,999) + 5(1+999) + 1(1+99) + 4(1+9) + 8
=6+3+5+1+4+8 + 6(99,999) + 3(9,999) + 5(999) + 1(99) + 4(9)
=(6+3+5+1+4+8) + 3[6(33,333) + 3(3,333) + 5(333) + 1(33) + 4(3)]
So, since the right hand part of the last line (the 3[6(33,333) + 3(3,333) + 5(333) + 1(33) + 4(3)] part) is divisible by 3, that means the whole number is divisible by 3 if and only if the left hand part of the last line (the (6+3+5+1+4+8) part) is also divisible by 3. Since this left hand part is simply the sum of the digits of the number, we get our divisibility test. Of course the same argument works to show that the divisibility by 9 test is analogous: an integer is divisible by 9 if and only if the sum of the digits is divisible by 9.

