`7^9+9^7` is divisible by (A) 16 (B) 24 (C) 64 (D) 72

Explanation

To determine whether \( 7^9 + 9^7 \) is divisible by any of the given options (16, 24, 64, or 72), we can use the Binomial Theorem and modular arithmetic. Here’s a step-by-step solution:

### Step 1: Rewrite the terms
We can express \( 7^9 \) and \( 9^7 \) in a more manageable form. Notice that:
\[ 7 = 8 - 1 \quad \text{and} \quad 9 = 8 + 1 \]
Thus, we can rewrite:
\[ 7^9 = (8 - 1)^9 \quad \text{and} \quad 9^7 = (8 + 1)^7 \]

### Step 2: Apply the Binomial Theorem
Using the Binomial Theorem, we expand both expressions:
\[ (8 - 1)^9 = \sum_{k=0}^{9} \binom{9}{k} 8^{9-k} (-1)^k \]
\[ (8 + 1)^7 = \sum_{j=0}^{7} \binom{7}{j} 8^{7-j} (1)^j \]

### Step 3: Combine the expansions
Now, we add these two expansions:
\[ 7^9 + 9^7 = (8 - 1)^9 + (8 + 1)^7 \]

### Step 4: Evaluate modulo 16
To check divisibility by 16, we can evaluate \( 7^9 \) and \( 9^7 \) modulo 16:
- \( 7 \equiv 7 \mod 16 \)
- \( 9 \equiv 9 \mod 16 \)

Calculating powers:
\[ 7^2 \equiv 49 \equiv 1 \mod 16 \quad \Rightarrow \quad 7^9 = (7^2)^4 \cdot 7 \equiv 1^4 \cdot 7 \equiv 7 \mod 16 \]
\[ 9^2 \equiv 81 \equiv 1 \mod 16 \quad \Rightarrow \quad 9^7 = (9^2)^3 \cdot 9 \equiv 1^3 \cdot 9 \equiv 9 \mod 16 \]

Adding these results:
\[ 7^9 + 9^7 \equiv 7 + 9 \equiv 16 \equiv 0 \mod 16 \]
Thus, \( 7^9 + 9^7 \) is divisible by 16.

### Step 5: Check divisibility by 24
Next, we check divisibility by 24:
- Check modulo 3:
- \( 7 \equiv 1 \mod 3 \quad \Rightarrow \quad 7^9 \equiv 1 \mod 3 \)
- \( 9 \equiv 0 \mod 3 \quad \Rightarrow \quad 9^7 \equiv 0 \mod 3 \)
- Thus, \( 7^9 + 9^7 \equiv 1 + 0 \equiv 1 \mod 3 \) (not divisible by 3)

Since \( 7^9 + 9^7 \) is not divisible by 3, it cannot be divisible by 24.

### Step 6: Check divisibility by 64
Now check divisibility by 64:
- \( 7 \equiv 7 \mod 64 \)
- \( 9 \equiv 9 \mod 64 \)

Calculating powers:
\[ 7^2 = 49 \quad \Rightarrow \quad 7^4 = 49^2 = 2401 \equiv 33 \mod 64 \]
\[ 7^8 = 33^2 = 1089 \equiv 1 \mod 64 \quad \Rightarrow \quad 7^9 \equiv 7 \mod 64 \]
\[ 9^2 = 81 \equiv 17 \mod 64 \quad \Rightarrow \quad 9^4 = 17^2 = 289 \equiv 33 \mod 64 \]
\[ 9^6 = 33 \cdot 17 = 561 \equiv 49 \mod 64 \quad \Rightarrow \quad 9^7 = 9 \cdot 49 = 441 \equiv 57 \mod 64 \]

Adding these results:
\[ 7^9 + 9^7 \equiv 7 + 57 \equiv 64 \equiv 0 \mod 64 \]
Thus, \( 7^9 + 9^7 \) is divisible by 64.

### Step 7: Check divisibility by 72
Check divisibility by 72:
- Check modulo 8:
- \( 7 \equiv -1 \mod 8 \quad \Rightarrow \quad 7^9 \equiv -1 \mod 8 \)
- \( 9 \equiv 1 \mod 8 \quad \Rightarrow \quad 9^7 \equiv 1 \mod 8 \)
- Thus, \( 7^9 + 9^7 \equiv -1 + 1 \equiv 0 \mod 8 \)

- Check modulo 9:
- \( 7 \equiv 7 \mod 9 \quad \Rightarrow \quad 7^9 \equiv 1 \mod 9 \)
- \( 9 \equiv 0 \mod 9 \quad \Rightarrow \quad 9^7 \equiv 0 \mod 9 \)
- Thus, \( 7^9 + 9^7 \equiv 1 + 0 \equiv 1 \mod 9 \) (not divisible by 9)

Since \( 7^9 + 9^7 \) is not divisible by 9, it cannot be divisible by 72.

### Conclusion
The only options that \( 7^9 + 9^7 \) is divisible by are 16 and 64. Since the question asks for one of the options, we can conclude that:

**The answer is (C) 64.**