AMC10 2024 B

$\textbf{First Case}$ We start off by knowing that there must exist an ordered pair (0,y,z) and a big term 7 such that the range is satisfied and the median could also be satisfied. Because x≤y≤z, we say x=0. Then, we get the sum 24.8. We need 24.8+y+z9∈Z. This means that we need the nearest multiple of 9, which is 36, and we get y+z=11.2. We need one of these to be the median, WLOG say y. Then, y∈{4,5}. So if y=4, we get z=7.2, and if y=5, we get 6.2. We see that if z=6.2, the range will still remain as 7, and therefore the one ordered triple (0,5,6.2) satisfies this. We know that the median can be y, so we have y∈{4,5}. We do casework on 4 and 5. $\textbf{Case 1}$ Say y=4. Then, because we already know there exists an ordered triple where z is not the largest, there must exist at least one ordered triple where z is, so we say z−x=7. We also know that the mean must be divisible by 9. Quickly summing each number up we get 24.8. The next number divisible by 9 is 36. We add y to get 28.8. We then know x+z=36−28.8=7.2. The smallest values give x=0.1 and z=7.1. This satisfies our constraints. We don't want any errors, so we check the next sum, which is 45. We get then 7.2=x+z and z−x=7. This gives x=4.6 and z=11.6. This is a contradiction, as x≤y≤z, and the values incrementally become too large, and therefore we only have one ordered triple for this case, which is (0.1,4,7.1) $\textbf{Case 2}$ We now check y=5. This makes our sum 29.8. The nearest multiple of 9 is 36, again. This gives us 6.2=x+z and z−x=7. Solving gives us -0.4 and 6.6. This however doesn't work, as 6.6 is not the greatest value. We decide to check the next multiple of 9, which is 45. This gives us 15.2=x+z and z−x=7. This gives us x=4.1 and z=11.1. This also doesn't work, as x becomes incrementally larger, and therefore there are no ordered triples that satisfy this. $\textbf{Last Case}$ We have one more case, and that is x is not the first number, but z is the last, meaning z−1=7, and z=8. This means that 24.8+x+y+z9∈Z. So, if z=8, we get 32.8. We again see that we need the nearest multiple of 9 which is 36, and we get 3.2=x+y. We see that x and y still need to be a integer median and 3.2 is too small, so 36 cannot work. We try 45, and see 12.2=x+y. After some trial and error, we see that x=6 and y=6.2, giving us the ordered triple (6,6.2,8). Continuing as before shows us that x again incrementally increases, and therefore we have only one ordered triple. We have $\boxed{\textbf{(C) }3}$ ordered triples. These are (0.1,4,7.1), (0,5,6.2), and (6,6.2,8).