If `9^7+7^9` is divisible b `2^n , `then find the greatest value of `n` ,where `n in N`.

Options

A. 5
B. 6
C. 7
D. None of these

Answer

To solve the problem of finding the greatest value of \( n \) such that \( 9^7 + 7^9 \) is divisible by \( 2^n \), we can follow these steps:

### Step 1: Rewrite the expression
We start with the expression \( 9^7 + 7^9 \). We can rewrite \( 9 \) as \( 8 + 1 \) and \( 7 \) as \( 8 - 1 \):
\[ 9^7 + 7^9 = (8 + 1)^7 + (8 - 1)^9 \]

### Step 2: Apply the Binomial Theorem
Using the Binomial Theorem, we expand both terms:
\[ (8 + 1)^7 = \sum_{k=0}^{7} \binom{7}{k} 8^k 1^{7-k} = 8^0 + 7 \cdot 8^1 + 21 \cdot 8^2 + 35 \cdot 8^3 + 35 \cdot 8^4 + 21 \cdot 8^5 + 7 \cdot 8^6 + 1 \cdot 8^7 \]
\[ (8 - 1)^9 = \sum_{k=0}^{9} \binom{9}{k} 8^k (-1)^{9-k} = -1 \cdot 8^0 + 9 \cdot 8^1 - 36 \cdot 8^2 + 84 \cdot 8^3 - 126 \cdot 8^4 + 126 \cdot 8^5 - 84 \cdot 8^6 + 36 \cdot 8^7 - 9 \cdot 8^8 + 1 \cdot 8^9 \]

### Step 3: Combine the expansions
Now we combine the two expansions:
\[ 9^7 + 7^9 = \left(1 + 7 \cdot 8 + 21 \cdot 8^2 + 35 \cdot 8^3 + 35 \cdot 8^4 + 21 \cdot 8^5 + 7 \cdot 8^6 + 8^7\right) + \left(-1 + 9 \cdot 8 - 36 \cdot 8^2 + 84 \cdot 8^3 - 126 \cdot 8^4 + 126 \cdot 8^5 - 84 \cdot 8^6 + 36 \cdot 8^7 - 9 \cdot 8^8 + 8^9\right) \]

### Step 4: Simplify the expression
When we simplify this expression, we notice that the constant terms cancel out:
\[ = (7 + 9) \cdot 8 + (21 - 36) \cdot 8^2 + (35 + 84) \cdot 8^3 + (35 - 126) \cdot 8^4 + (21 + 126) \cdot 8^5 + (7 - 84) \cdot 8^6 + (1 + 36) \cdot 8^7 + (8^9 - 9 \cdot 8^8) \]

### Step 5: Factor out powers of 2
Notice that \( 8 = 2^3 \), so we can factor out \( 2^6 \) from the expression:
\[ = 2^6 \cdot (k) \quad \text{(where \( k \) is some integer)} \]

### Step 6: Determine the value of \( n \)
Since we have factored out \( 2^6 \), we conclude that \( 9^7 + 7^9 \) is divisible by \( 2^6 \). Therefore, the greatest value of \( n \) such that \( 9^7 + 7^9 \) is divisible by \( 2^n \) is:
\[ \boxed{6} \]