`(32^34)^35` divided by 7, the reamainder obtained is ?

Options

A. 5
B. 4
C. 6
D. 2

Answer

To find the remainder of \((32^{34})^{35}\) when divided by 7, we can simplify the expression step by step.

### Step 1: Simplify the base
First, we can simplify \(32\) modulo \(7\):
\[ 32 \mod 7 = 4 \]
So, we can rewrite the expression:
\[ (32^{34})^{35} \equiv (4^{34})^{35} \mod 7 \]

### Step 2: Simplify the exponent
Next, we can simplify the exponent:
\[ (4^{34})^{35} = 4^{34 \times 35} = 4^{1190} \]

### Step 3: Use Fermat's Little Theorem
According to Fermat's Little Theorem, if \(p\) is a prime number and \(a\) is an integer not divisible by \(p\), then:
\[ a^{p-1} \equiv 1 \mod p \]
Here, \(p = 7\) and \(a = 4\). Since \(4\) is not divisible by \(7\), we can apply the theorem:
\[ 4^{6} \equiv 1 \mod 7 \]

### Step 4: Reduce the exponent modulo \(6\)
Now, we need to reduce \(1190\) modulo \(6\):
\[ 1190 \mod 6 \]
Calculating \(1190\) divided by \(6\):
\[ 1190 \div 6 = 198 \quad \text{(remainder } 2\text{)} \]
So, \(1190 \equiv 2 \mod 6\).

### Step 5: Substitute back into the expression
Now we can substitute back into our expression:
\[ 4^{1190} \equiv 4^{2} \mod 7 \]

### Step 6: Calculate \(4^{2} \mod 7\)
Now we calculate \(4^{2}\):
\[ 4^{2} = 16 \]
Now we find \(16\) modulo \(7\):
\[ 16 \mod 7 = 2 \]

### Final Answer
Thus, the remainder when \((32^{34})^{35}\) is divided by \(7\) is:
\[ \boxed{2} \]