`(32^32 )^32` is divided by 7 , the remainder obtained is ?

Options

A. a) 4
B. b) 3
C. c) 5
D. d) 2

Answer

To find the remainder when \((32^{32})^{32}\) is divided by 7, we can follow these steps:

### Step 1: Simplify the Base
First, we note that \(32\) can be expressed in terms of a smaller base modulo \(7\):
\[ 32 \mod 7 = 4 \]
So, we can rewrite the expression as:
\[ (32^{32})^{32} \equiv (4^{32})^{32} \mod 7 \]

### Step 2: Simplify the Exponent
Now we simplify the exponent:
\[ (4^{32})^{32} = 4^{32 \times 32} = 4^{1024} \]

### 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:
\[ a^{p-1} \equiv 1 \mod p \]
In our case, \(p = 7\) and \(a = 4\). Thus:
\[ 4^{6} \equiv 1 \mod 7 \]

### Step 4: Reduce the Exponent Modulo 6
Now we need to reduce the exponent \(1024\) modulo \(6\) (since \(6 = 7 - 1\)):
\[ 1024 \mod 6 \]
Calculating \(1024 \div 6\):
\[ 1024 = 6 \times 170 + 4 \]
So,
\[ 1024 \mod 6 = 4 \]

### Step 5: Calculate \(4^4 \mod 7\)
Now we can compute:
\[ 4^{1024} \equiv 4^{4} \mod 7 \]
Calculating \(4^4\):
\[ 4^2 = 16 \]
Now, reduce \(16\) modulo \(7\):
\[ 16 \mod 7 = 2 \quad \text{(since } 16 - 14 = 2\text{)} \]
Now calculate \(4^4\):
\[ 4^4 = (4^2)^2 = 16^2 = 256 \]
Now reduce \(256\) modulo \(7\):
\[ 256 \mod 7 \]
Calculating \(256 \div 7\):
\[ 256 = 7 \times 36 + 4 \]
So,
\[ 256 \mod 7 = 4 \]

### Final Step: Conclusion
Thus, the remainder when \((32^{32})^{32}\) is divided by \(7\) is:
\[ \boxed{4} \]