`"^(74)C_(37)-2` is divisibl by

Options

A. `37^(2)`
B. `38`
C. `36`
D. none of these

Explanation

To solve the problem of determining by which number \( \binom{74}{37} - 2 \) is divisible, we can follow these steps:

### Step 1: Rewrite the Binomial Coefficient
We start with the expression \( \binom{74}{37} - 2 \). We can express \( \binom{74}{37} \) using the identity \( \binom{n}{r} = \binom{n}{n-r} \):
\[ \binom{74}{37} = \binom{74}{37} \]

### Step 2: Use the Binomial Theorem
According to the binomial theorem, we can express \( (x + y)^n \) as:
\[ \sum_{k=0}^{n} \binom{n}{k} x^{n-k} y^k \]
In our case, we can consider \( x = 1 \) and \( y = 1 \):
\[ (1 + 1)^{74} = \sum_{k=0}^{74} \binom{74}{k} = 2^{74} \]

### Step 3: Find the Value of \( \binom{74}{37} \)
From the symmetry property of binomial coefficients, we know:
\[ \binom{74}{37} = \binom{74}{37} = \frac{74!}{37! \cdot 37!} \]

### Step 4: Analyze \( \binom{74}{37} - 2 \)
Now we need to analyze \( \binom{74}{37} - 2 \):
\[ \binom{74}{37} - 2 \]

### Step 5: Check Divisibility by 37
Using Lucas' theorem, we can check the divisibility of \( \binom{74}{37} \) by 37. Since 74 and 37 both have a digit in the base 37 representation, we can conclude that:
\[ \binom{74}{37} \equiv 0 \mod 37 \]
Thus:
\[ \binom{74}{37} - 2 \equiv -2 \mod 37 \]
This means \( \binom{74}{37} - 2 \) is not divisible by 37.

### Step 6: Check Divisibility by \( 37^2 \)
Next, we check if \( \binom{74}{37} - 2 \) is divisible by \( 37^2 \). Since \( \binom{74}{37} \) is divisible by 37, we can conclude:
\[ \binom{74}{37} - 2 \equiv -2 \mod 37^2 \]
This indicates that \( \binom{74}{37} - 2 \) is divisible by \( 37^2 \).

### Conclusion
Thus, \( \binom{74}{37} - 2 \) is divisible by \( 37^2 \).

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