What is the smallest number greater than 1000 that when divided by any one of the numbers 6, 9, 12, 15, 18 leaves a remainder of 3?

To find the smallest number greater than 1000 that leaves a remainder of 3 when divided by 6, 9, 12, 15, 18, we first need to find the Least Common Multiple (LCM) of these divisors.

1. Calculate the LCM of 6, 9, 12, 15, 18: Prime factorization: 6 = 2 x 3 9 = 3^2 12 = 2^2 x 3 15 = 3 x 5 18 = 2 x 3^2 LCM = 2^2 x 3^2 x 5 = 4 x 9 x 5 = 180.

2. Any number that leaves a remainder of 3 when divided by 6, 9, 12, 15, 18 must be of the form (LCM * k) + 3, where k is an integer. So, the number is of the form 180k + 3.

3. We need the smallest such number that is greater than 1000. 180k + 3 > 1000 180k > 997 k > 997 / 180 k > 5.538...

4. The smallest integer value for k that satisfies k > 5.538... is k = 6.

5. Substitute k = 6 into the expression: Number = 180 * 6 + 3 Number = 1080 + 3 Number = 1083.

Analysis of Options: A) 1063: 1063 - 3 = 1060. 1060 is not divisible by 180 (1060/180 is not an integer). B) 1073: 1073 - 3 = 1070. 1070 is not divisible by 180. C) 1083: 1083 - 3 = 1080. 1080 is divisible by 180 (1080 / 180 = 6). This is the correct number. D) 1183: 1183 - 3 = 1180. 1180 is not divisible by 180.

Therefore, 1083 is the smallest number greater than 1000 that satisfies the given conditions.

The final answer is C