When 391758 and 394915 are divided by a certain three digit number, the three digit remainder is the same in each case. Find the divisor. Let D = the required divisor Let Q1 = quotient when 391758 is divided by D Let Q2 = quotient when 394915 is divided by D Let R - the remainder in each case Then 391758 = Q1·D + R 394915 = Q2·D + R Subtracting the 1st equation from the 2nd 3157 = Q1·D - Q2·D 3157 = D(Q2 - Q1) D = 3157/(Q2 - Q1) So Q2 - Q1 has to be a factor of 3157 3157 = 7×11×41 So the factors of 3157 are 1, 7, 11, 7×11, 7×41, 11×41, and 7×11×41 So Q2-Q1 is one of these: 1, 7, 11, 77, 287, 451, and 3157 So D = 3157, 451, 287, 41, 11, 7, or 1 Only two of these have three digits, 287 and 451. When 391758 and 394915 are divided by 287, the remainder is 3 in both cases. But the remainder must be a 3-digit number When 391758 and 394915 are divided by 451, the remainder is 290 in both cases. That's a 3-digit remainder. so 451 is the answer. Edwin AnlytcPhil@aol.com