To find the least possible number with which 270! can be exactly divisible by (54)48, we need to understand the concept of factorials, divisibility rules, and prime factorization.

Factorials:
A factorial is the product of an integer and all the positive integers below it. For example, 5! (read as "5 factorial") is calculated as 5 × 4 × 3 × 2 × 1 = 120.

Divisibility Rules:
To determine if a number is divisible by another number, we need to check certain divisibility rules. The divisibility rule for 48 is that a number must be divisible by both 4 and 12.

Prime Factorization:
Prime factorization is the process of finding the prime numbers that multiply together to create a given number. We can express a number as a product of its prime factors. For example, prime factorization of 48 is 2 × 2 × 2 × 2 × 3.

Now let's calculate the least possible number with which 270! can be exactly divisible by (54)48.

Step 1: Prime Factorization of (54)48
(54)48 = 54 × 48

Prime factorization of 54:
54 = 2 × 3 × 3 × 3
= 2 × 3^3

Prime factorization of 48:
48 = 2 × 2 × 2 × 2 × 3
= 2^4 × 3

Now we have the prime factorization of (54)48 as:
(54)48 = (2 × 3^3) × (2^4 × 3)
= 2^5 × 3^4

Step 2: Prime Factorization of 270!
To find the least possible number, we need to determine the highest powers of 2 and 3 in the prime factorization of 270!.

Prime factorization of 270! can be calculated by finding the highest power of 2 and 3 in the multiples of 2 and 3 respectively.

Highest power of 2 in 270!:
270 ÷ 2 = 135
135 ÷ 2 = 67.5 (ignore decimal, consider only the quotient)
67 ÷ 2 = 33.5
33 ÷ 2 = 16.5
16 ÷ 2 = 8
8 ÷ 2 = 4
4 ÷ 2 = 2
2 ÷ 2 = 1

Adding all the quotients: 135 + 67 + 33 + 16 + 8 + 4 + 2 + 1 = 266

Therefore, the highest power of 2 in 270! is 2^266.

Highest power of 3 in 270!:
270 ÷ 3 = 90
90 ÷ 3 = 30
30 ÷ 3 = 10
10 ÷ 3 = 3.3 (ignore decimal, consider only the quotient)
3 ÷ 3 = 1

Adding all the quotients: 90 + 30 + 10 + 3 + 1 = 134

Therefore, the highest power of 3 in