(3) Here, difference between divisorand remainder is same in bothcases = (18 – 4) = (30 – 16) = 14 ∴ Required number = k × LCM of(18, 30) - Common difference Where k is 1, 2, 3 … LCM of 18 and 30 is 90. Largest three digit multiple of 90 =990 ∴ Maximum possible number ofchocolates = 990 – 14 = 976