The least possible number with which (180)! should be multiplied so that it can be divided by `(45)^(44)` is:

Options

A. 44
B. 25
C. 9
D. none of these

Answer

To solve the problem of finding the least possible number with which \( (180)! \) should be multiplied so that it can be divided by \( (45)^{44} \), we will follow these steps:

### Step 1: Factorize \( 45 \)
First, we need to factor \( 45 \):
\[ 45 = 5 \times 9 = 5 \times 3^2 \]
Thus, we can express \( (45)^{44} \) as:
\[ (45)^{44} = (5 \times 3^2)^{44} = 5^{44} \times 3^{88} \]

### Step 2: Determine the required powers of 3 and 5
From the factorization, we see that we need \( (180)! \) to have at least \( 5^{44} \) and \( 3^{88} \).

### Step 3: Calculate the number of factors of 5 in \( (180)! \)
To find the number of factors of \( 5 \) in \( (180)! \), we use the formula:
\[ \text{Number of factors of } p = \left\lfloor \frac{n}{p} \right\rfloor + \left\lfloor \frac{n}{p^2} \right\rfloor + \left\lfloor \frac{n}{p^3} \right\rfloor + \ldots \]
For \( p = 5 \) and \( n = 180 \):
\[ \left\lfloor \frac{180}{5} \right\rfloor + \left\lfloor \frac{180}{25} \right\rfloor + \left\lfloor \frac{180}{125} \right\rfloor \]
Calculating each term:
\[ \left\lfloor \frac{180}{5} \right\rfloor = 36, \quad \left\lfloor \frac{180}{25} \right\rfloor = 7, \quad \left\lfloor \frac{180}{125} \right\rfloor = 1 \]
Adding these:
\[ 36 + 7 + 1 = 44 \]
So, \( (180)! \) has exactly \( 44 \) factors of \( 5 \).

### Step 4: Calculate the number of factors of 3 in \( (180)! \)
Now, we calculate the number of factors of \( 3 \) in \( (180)! \):
\[ \left\lfloor \frac{180}{3} \right\rfloor + \left\lfloor \frac{180}{9} \right\rfloor + \left\lfloor \frac{180}{27} \right\rfloor + \left\lfloor \frac{180}{81} \right\rfloor \]
Calculating each term:
\[ \left\lfloor \frac{180}{3} \right\rfloor = 60, \quad \left\lfloor \frac{180}{9} \right\rfloor = 20, \quad \left\lfloor \frac{180}{27} \right\rfloor = 6, \quad \left\lfloor \frac{180}{81} \right\rfloor = 2 \]
Adding these:
\[ 60 + 20 + 6 + 2 = 88 \]
So, \( (180)! \) has exactly \( 88 \) factors of \( 3 \).

### Step 5: Determine if additional factors are needed
We need \( (180)! \) to have at least \( 5^{44} \) and \( 3^{88} \):
- For \( 5^{44} \): \( (180)! \) already has \( 44 \) factors of \( 5 \), so no additional factors are needed.
- For \( 3^{88} \): \( (180)! \) already has \( 88 \) factors of \( 3 \), so no additional factors are needed.

### Conclusion
Since \( (180)! \) already contains the required factors of both \( 5 \) and \( 3 \), the least possible number with which \( (180)! \) should be multiplied is:
\[ \boxed{1} \]