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Hi Tonkotsu
We are keeping it constant to arrive at a benchmark for the sum of all terms. We can see that the minimum sum of all terms will be 90 and the maximum sum will be 180. Also, we are given that sum of all terms is 104.

Sum if all the terms are constant:
3 = 90
4 = 120
5 = 150
6 = 180

104 is between 90 & 120. The distance of 104 from 90 is 14, and from 120 is 16. So there must be more number of 3's than 4's (this is the least possible scenario).
Lets us assume that there is a 5/6 in the set, so to balance the total sum we will have to increase the number of 3's in the set , but the median will remain same as we already have established that the 15th & 16th terms will be 3 only.
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