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?

Correct Answer: (C) 1083

The required number must be of the form $\text{LCM}(6,9,12,15,18)\times k+3$$\text{LCM}$(6, 9, 12, 15, 18) $\times$ k + 3. Calculating the LCM: $6 = 2$$\times$ 3, $9 = 3^2$, $12 = 2^2$$\times$ 3, $15 = 3$$\times$ 5, and $18 = 2$$\times$ 3^2, so $\text{LCM}=2^2\times 3^2\times 5=180$$\text{LCM}$ = 2^2 $\times$ 3^2 $\times$ 5 = 180. Seeking the smallest such number strictly greater than 1000: For $k = 5$, $180$$\times$ 5 + 3 = 903 $\le$ 1000. For $k = 6$, $180$$\times$ 6 + 3 = 1080 + 3 = 1083 > 1000. Thus, 1083 is the smallest valid number.