2019 Mock AMC 10B Problems/Problem 16
Solution by excruciating: For each of , they all need to be at least , so we have: or without the " need to be restriction." So now it's stars and bars:
which simplifies to which we can write as .
Expanding, we get .
Because we only have terms, is to be around the highest divisor. Thus, we must have as a divisor too. is one too. We have left is and , to make terms, so we have , and . Thus we have:
Thus,
Note: does not need to be because the inequality is not strict. So this solution is incorrect.