Calvin's Calorie intake - (2)

From the conditions , we are looking for the smallest number above $1000$ that is a multiple of $6 \times 7 = 42$ . Since $\lceil\frac{1000}{42}\rceil \approx \lceil 23.8 \rceil = 24$ , the maximum number of calories that Calvin could have consumed is $24 \times 42 = \boxed{1008}$

Since $6*7=42$ , we are looking for the smallest multiple of $42$ above $1000$ . This will equal $\lceil1000/42\rceil=24$ . Therefore, the answer is $42\times24=1008$ .