Problem: Find the greatest 4-digit number which when divided by 12, 18, 36, and 54 leaves a remainder of 8 in each case.

Solution:

To solve this problem, we need to find a number that satisfies the following conditions:
- The number is greater than or equal to 1000 and less than or equal to 9999.
- When the number is divided by 12, 18, 36, and 54, it leaves a remainder of 8 in each case.

We can solve this problem using the Chinese Remainder Theorem (CRT).

Step 1: Find the LCM of 12, 18, 36, and 54.

LCM(12, 18, 36, 54) = 324

Step 2: Express the conditions in terms of the LCM.

Let N be the number we are looking for. Then, we can write:

N ≡ 8 (mod 12)
N ≡ 8 (mod 18)
N ≡ 8 (mod 36)
N ≡ 8 (mod 54)

Using the LCM, we can rewrite these conditions as follows:

N ≡ 8 (mod 324)

Step 3: Find the largest number that satisfies the conditions.

To find the largest number that satisfies the conditions, we need to subtract 8 from the LCM and check if the result is less than or equal to 9999.

324 - 8 = 316

So, we need to check if 316 is less than or equal to 9999.

Since 316 is less than 1000, we need to add the LCM to it until we get a number that is greater than or equal to 1000.

316 + 324 = 640
640 + 324 = 964
964 + 324 = 1288

Now, we need to check if 1288 satisfies the conditions.

1288 ≡ 8 (mod 12)
1288 ≡ 8 (mod 18)
1288 ≡ 8 (mod 36)
1288 ≡ 8 (mod 54)

So, 1288 is the largest 4-digit number that satisfies the conditions.

Answer: The greatest 4-digit number that when divided by 12, 18, 36, and 54 leaves a remainder of 8 in each case is 1288.