Let the required number be x and the common remainder be r, so we can write 82 as ax plus r, 99 as bx plus r, and 150 as cx plus r. Subtracting the equations pairwise gives 17, 51, and 68, meaning x must perfectly divide the HCF of these differences. The HCF of 17, 51, and 68 is 17, which when tested with the original numbers leaves a common remainder of 14. The greatest such number is 17.