Using clues 1 and 2 together, we can determine that the carbs in P2 must be less than 75 but at least greater
than 62.
The only possible values are 65 and 70
Putting 65 grams of carbs in P2 gives us 100-65-14-8 = 13 grams of protein, which is invalid.
Putting 70 grams of carbs in P2 gives 100-70-14-8 = 8 grams of protein.
Clue 1 gives us that the protein in M2 should be less than that present in either of the pseudo-cereals, which we
right now have a baseline of 14.
So the protein in M2 (and M3) can be 12, 8, 4, 0

The protein and carbs in M2 should add up to 100-7-16 = 77
This is only possible if the protein count is 12, giving carbs as 65
The protein in M3 can be 0, 4, 8 or 12
We are given in clue five that the protein in P1 is double that in M3.
The protein in P1 thus can be 0, 8, 16 or 24
Since this P1 protein also has to be more than M1 and M2 protein, it can not be 0 or 8
Leaving only 16 or 24 as the valid values.
The protein and fats in P1 must add up to 100-66-10 = 24
If P1 had 24 grams of protein, then it would have 0 grams of fat, but in clue 4, we are given that all missing fats
are non-zero multiples of 4
Hence, the only possible protein value in P1 is 16, with 8 grams of fats. Giving 8 grams of protein in M3 and 24
grams of others in M3.
We can see that there were 12 grams of protein in M2.