When `(32^32)^32` is divided by 9, the remainder obtained is ?

Options

A. 4
B. 7
C. 2
D. 1

Answer

To find the remainder when \((32^{32})^{32}\) is divided by 9, we can simplify the expression step by step.

### Step 1: Simplify the Expression
We start with the expression \((32^{32})^{32}\). By the laws of exponents, we can simplify this to:
\[ 32^{32 \times 32} = 32^{1024} \]

### Step 2: Find the Remainder of 32 Divided by 9
Next, we need to find the remainder when 32 is divided by 9. We perform the division:
\[ 32 \div 9 = 3 \quad \text{(since } 9 \times 3 = 27\text{)} \]
The remainder is:
\[ 32 - 27 = 5 \]
So, we have:
\[ 32 \equiv 5 \mod 9 \]

### Step 3: Substitute Back into the Expression
Now we can substitute \(32\) with \(5\) in our expression:
\[ 32^{1024} \equiv 5^{1024} \mod 9 \]

### Step 4: Use Fermat's Little Theorem
Since \(5\) and \(9\) are coprime, we can use Fermat's Little Theorem, which states that if \(p\) is a prime and \(a\) is an integer not divisible by \(p\), then:
\[ a^{p-1} \equiv 1 \mod p \]
Here, \(p = 9\) and \(a = 5\). Thus:
\[ 5^{6} \equiv 1 \mod 9 \quad \text{(since } 9 - 1 = 8 \text{ and } 8 \text{ is not prime, we use } 6 \text{ for } 9\text{)} \]

### Step 5: Reduce the Exponent Modulo 6
Now, we need to reduce \(1024\) modulo \(6\):
\[ 1024 \div 6 = 170 \quad \text{(with a remainder of } 4\text{)} \]
This means:
\[ 1024 \equiv 4 \mod 6 \]

### Step 6: Calculate \(5^{1024} \mod 9\)
Now we can compute:
\[ 5^{1024} \equiv 5^{4} \mod 9 \]
Calculating \(5^{4}\):
\[ 5^{2} = 25 \]
Now find \(25 \mod 9\):
\[ 25 \div 9 = 2 \quad \text{(remainder } 7\text{)} \]
So,
\[ 5^{2} \equiv 7 \mod 9 \]
Now calculate \(5^{4}\):
\[ 5^{4} = (5^{2})^{2} \equiv 7^{2} \mod 9 \]
Calculating \(7^{2}\):
\[ 7^{2} = 49 \]
Now find \(49 \mod 9\):
\[ 49 \div 9 = 5 \quad \text{(remainder } 4\text{)} \]
Thus,
\[ 5^{4} \equiv 4 \mod 9 \]

### Conclusion
The remainder when \((32^{32})^{32}\) is divided by \(9\) is:
\[ \boxed{4} \]