Which of the following number does not completely divide `(3^41 + 7^82)` ?

Options

A. a) 4
B. b) 52
C. c) 17
D. d) 26

Answer

To determine which of the following numbers does not completely divide \(3^{41} + 7^{82}\), we can follow these steps:

### Step 1: Rewrite the Expression
We start with the expression:
\[ 3^{41} + 7^{82} \]
We can rewrite \(7^{82}\) as \((7^2)^{41} = 49^{41}\). Therefore, we have:
\[ 3^{41} + 49^{41} \]

### Step 2: Apply the Sum of Powers Rule
According to the mathematical rule for the sum of powers, if we have \(P^n + Q^n\) where \(n\) is odd, then \(P^n + Q^n\) is divisible by \(P + Q\). Here, \(n = 41\) (which is odd), \(P = 3\), and \(Q = 49\). Thus:
\[ 3^{41} + 49^{41} \text{ is divisible by } 3 + 49 = 52 \]

### Step 3: Check the Divisibility of 52
Next, we need to consider the factors of 52. The factors of 52 are:
- \(1\)
- \(2\)
- \(4\)
- \(13\)
- \(26\)
- \(52\)

### Step 4: Identify Which Number Does Not Divide
Now we need to check which of the given options does not divide \(3^{41} + 7^{82}\). From our previous analysis, we found that \(3^{41} + 49^{41}\) is divisible by 52.

Next, we check the divisibility of 52 by its factors:
- \(52\) is divisible by \(1\), \(2\), \(4\), \(13\), and \(26\).
- However, we need to check if \(17\) divides \(3^{41} + 7^{82}\).

### Step 5: Check Divisibility by 17
To check if \(3^{41} + 7^{82}\) is divisible by \(17\), we can use Fermat's Little Theorem, which states that if \(p\) is a prime number and \(a\) is not divisible by \(p\), then:
\[ a^{p-1} \equiv 1 \mod p \]
For \(p = 17\):
- \(3^{16} \equiv 1 \mod 17\)
- \(7^{16} \equiv 1 \mod 17\)

Now, we can reduce the exponents:
- \(41 \mod 16 = 9\) (since \(41 = 2 \times 16 + 9\))
- \(82 \mod 16 = 2\) (since \(82 = 5 \times 16 + 2\))

Now we calculate:
\[ 3^{41} \equiv 3^9 \mod 17 \]
Calculating \(3^9\):
- \(3^2 = 9\)
- \(3^4 = 9^2 = 81 \equiv 13 \mod 17\)
- \(3^8 = 13^2 = 169 \equiv 16 \mod 17\)
- \(3^9 = 3^8 \cdot 3 \equiv 16 \cdot 3 = 48 \equiv 14 \mod 17\)

Now for \(7^{82}\):
\[ 7^{82} \equiv 7^2 \mod 17 \]
Calculating \(7^2\):
\[ 7^2 = 49 \equiv 15 \mod 17 \]

Now we combine:
\[ 3^{41} + 7^{82} \equiv 14 + 15 = 29 \equiv 12 \mod 17 \]

Since \(12 \not\equiv 0 \mod 17\), \(3^{41} + 7^{82}\) is not divisible by \(17\).

### Conclusion
Thus, the number that does not completely divide \(3^{41} + 7^{82}\) is:
\[ \boxed{17} \]