What is the greatest four-digit number that leaves a remainder of 4 when divided by 6, 9, or 12?

The correct option is 9976.

Step-by-step explanation:

Let the required number be N.

According to the question, when N is divided by 6, 9, or 12, it leaves a remainder of 4 in each case.
This means that (N - 4) is exactly divisible by 6, 9, and 12.

Step 1: Find the Least Common Multiple (LCM) of 6, 9, and 12

Prime factorizations of the numbers are:
6 = 2 × 3
9 = 3²
12 = 2² × 3

$\text{LCM}(6,9,12)=2^2\times 3^2=4\times 9=36$

Therefore, any number that is divisible by 6, 9, and 12 must be a multiple of 36.

Step 2: Find the greatest 4-digit number divisible by 36

The greatest 4-digit number is 9999.

Now, divide 9999 by 36 to find the remainder:
9999 ÷ 36 = 277 with a remainder of 27.

Subtract the remainder from 9999 to get the greatest 4-digit multiple of 36:
9999 - 27 = 9972.

So, 9972 is the greatest 4-digit number that is exactly divisible by 6, 9, and 12.

Step 3: Add the required remainder

Since the required number N leaves a remainder of 4 when divided by 6, 9, or 12, we add 4 to 9972:

$N=9972+4=9976$

Thus, the greatest four-digit number that leaves a remainder of 4 when divided by 6, 9, or 12 is 9976.