 So, let's put all of these procedures to use and find a number that is 2 more than the multiple of 20, 2 more than the multiple of 15, and 6 more than the multiple of 28. So, first, we'll reduce our divisors. Since 20 and 28 have a common factor, we divide the larger by the common factor and replace, so 28 divided by 4 gives us 7. Since 20 and 15 have a common factor, we divide the larger by the common factor and replace 20 divided by 5 is equal to 4. So now we have a new set of divisors. The remainders for now remain the same, and our new problem when counted by 4s to remain, when counted by 15s to remain, and when counted by 7s, 6s remain. Now we find our set numbers. So first, we want to find a multiple of 15 times 7, 105, that's 1 more than the multiple of 4. So applying Kinje Shao's procedure, we begin by noting that 1 105 is 105 more than the multiple of 4, namely 0 times 4, and 0 105s is 4 less than the multiple of 4, namely 1 times 4. We'll apply Kinje Shao's procedure, so we repeatedly subtract the smaller number from the larger number, and we can do that 26 times, multiply the corner numbers and add, and we can read our result as saying that 1 105 is 1 more than the multiple of 4, and so for every unit left when dividing by 4, set 105. So next, we want to find a multiple of 7 times 4, 28, that's 1 more than the multiple of 15. So 1 28 is 28 more than the multiple of 15, namely 0 times 15, and 0 28 is 15 less. Our first run through Kinje Shao's procedure tells us 1 28 is 13 more than the multiple of 15. If we apply Kinje Shao's procedure again, 1 28 is 2 less, then 7 28 is 1 more, and even though we don't need this, 8 28 is 1 less, and so that gives us our second set number for every unit left when dividing by 15, set 7 times 28, 196. And our third set number, we want to find a multiple of 4 times 15, 60, that's 1 more than the multiple of 7. So 1 60 is 60 more, 0 60s is 7 less. Applying our procedure, and so we read our last table, 2 60s is 1 more than the multiple of 7, so for every unit remaining when dividing by 7, we set down to 60s, 120. So to recap, our original problem, we reduced our divisors, and then we found the set numbers for every unit left when dividing by 4, set 105, for every unit left when dividing by 15, set 196, and for every unit left when dividing by 7, set 120. And so we can get a solution by setting down 2 105s, 2 196s, and 6 120s, which gives us 1 solution, and we can reduce this by 4 times 15 times 7, 420, to get smaller solutions, and the smallest solution is 62.