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

The correct option is 1083.

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

Step 1: Find the LCM of 6, 9, 12, 15, and 18
We first find the prime factorization of each number:
6 = 2 × 3
9 = 3²
12 = 2² × 3
15 = 3 × 5
18 = 2 × 3²

To find the LCM, we take the highest power of each prime factor that appears in these factorizations:
- The highest power of 2 is 2².
- The highest power of 3 is 3².
- The highest power of 5 is 5¹.

$\mathrm{LCM}(6,9,12,15,18)=2^2\times 3^2\times 5=4\times 9\times 5=180$

Thus, any number that is perfectly divisible by all these five numbers must be a multiple of 180.

Step 2: Write the general formula for the required number
Since the number must leave a remainder of 3 when divided by these numbers, it must be of the form:

$N=180k+3$

where k is a positive integer.

Step 3: Find the smallest value of k such that N is greater than 1000
Let us test different integer values for k to find the first one that makes N > 1000:
For k = 5:

$N=180\times 5+3=900+3=903$

Since 903 is less than 1000, we check the next integer, k = 6:

$N=180\times 6+3=1080+3=1083$

Since 1083 is greater than 1000, it is the smallest number that satisfies all the given conditions.