What is the smallest 4-digit number which, when divided by 12, 16, and 20, gives a remainder of 9?

Correct option is A

Given:

We need to find the smallest 4-digit number that, when divided by 12, 16, and 20, leaves a remainder of 9.

Concept Used:

The Least Common Multiple (LCM) of 12, 16, and 20, and then find the smallest 4-digit number that is 9 more than a multiple of this LCM.

Solution:

Prime factorization:

12 = 2² × 3

16 = 2⁴

20 = 2² × 5

LCM = 2⁴ × 3 × 5 = 240

The smallest 4-digit number is 1000.

So,

​$$240k + 9 \geq 1000 \\ \ \\240k \geq 991\\ \ \\k \geq \frac{991}{240} \approx 4.129$$​​

Since k must be an integer, the smallest k is 5.

So, the number:

$$240 \times 5$$​ + 9 = 1200 + 9 = 1209