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. Find 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. The numbers that leave a remainder of 3 when divided by 6, 9, 12, 15, 18 are of the form (LCM * k) + 3, where k is a positive integer. So, the numbers are of the form 180k + 3.

3. We need the smallest such number that is greater than 1000. Set 180k + 3 > 1000 180k > 997 k > 997 / 180 k > 5.53...

4. Since k must be an integer, the smallest integer value for k that satisfies this condition is k = 6.

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

6. Checking the options: A) 1063: (1063 - 3) = 1060. 1060 is not divisible by 180. 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 fits the pattern and is the smallest number greater than 1000. D) 1183: (1183 - 3) = 1180. 1180 is not divisible by 180.

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

The final answer is C