Suppose ,m divided by n , then quotient q and remainder r `{:("n)m(q"),(" "-), (" "-), (" "r) , (" "):}` or ` m= nq + r , AA m,n,q, r in 1 and n ne 0 ` If a is the remainder when ` 5^(40)` us divided by 11 and b is the remainder when ` 2^(2011)` is divided by 17 , the value of a + b is

Options

A. 7
B. 8
C. 9
D. 10

Explanation

To solve the problem, we need to find the remainders \( a \) and \( b \) as described in the question, and then add them together.

### Step 1: Calculate \( a \) (the remainder when \( 5^{40} \) is divided by \( 11 \))

We can use Fermat's Little Theorem, which states that 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, \( p = 11 \) and \( a = 5 \). Since \( 5 \) is not divisible by \( 11 \), we can apply the theorem:
\[ 5^{10} \equiv 1 \mod 11 \]
Now, we need to express \( 5^{40} \) in terms of \( 5^{10} \):
\[ 5^{40} = (5^{10})^4 \equiv 1^4 \equiv 1 \mod 11 \]
Thus, the remainder \( a \) is:
\[ a = 1 \]

### Step 2: Calculate \( b \) (the remainder when \( 2^{2011} \) is divided by \( 17 \))

Again, we can use Fermat's Little Theorem. Here, \( p = 17 \) and \( a = 2 \). Since \( 2 \) is not divisible by \( 17 \):
\[ 2^{16} \equiv 1 \mod 17 \]
Next, we need to find \( 2011 \mod 16 \):
\[ 2011 \div 16 = 125 \quad \text{(quotient)} \]
\[ 2011 - (125 \times 16) = 11 \quad \text{(remainder)} \]
Thus, \( 2011 \equiv 11 \mod 16 \). Therefore:
\[ 2^{2011} \equiv 2^{11} \mod 17 \]
Now we calculate \( 2^{11} \):
\[ 2^1 = 2 \]
\[ 2^2 = 4 \]
\[ 2^3 = 8 \]
\[ 2^4 = 16 \equiv -1 \mod 17 \]
\[ 2^5 = 32 \equiv 15 \mod 17 \]
\[ 2^6 = 64 \equiv 13 \mod 17 \]
\[ 2^7 = 128 \equiv 11 \mod 17 \]
\[ 2^8 = 256 \equiv 7 \mod 17 \]
\[ 2^9 = 512 \equiv 14 \mod 17 \]
\[ 2^{10} = 1024 \equiv 11 \mod 17 \]
\[ 2^{11} = 2048 \equiv 8 \mod 17 \]
Thus, the remainder \( b \) is:
\[ b = 8 \]

### Step 3: Calculate \( a + b \)

Now we can find \( a + b \):
\[ a + b = 1 + 8 = 9 \]

### Final Answer
The value of \( a + b \) is \( 9 \).

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