Question 1161162
The second digit is either a 0 or a 5
Therefore, need a multiplier of 3 and 7 that ends in 9 or 4. There are no suitable ones for 9 that are evenly divisible by both 3 and 7
for 4, there is the product (3*7)*4 or 84
Then there would be 85 eggs, which satisfies the second part. ANSWER
Removing them 3 at a time, 84 could be removed
same for 7 at a time.  No other product of 3 and 7 ends in four for both when < 100