Question

If $2^{4n+3} + 3^{3n+1}$ is divisible by P for all natural numbers $n$, then P is

• A. an even integer
• B. an odd integer, not a prime
• C. an odd prime integer
• D. an integer less than 9

Hint:

To find a number that divides an expression for all natural $n$, evaluate the expression for a few initial values of $n$ and determine the common factor (GCD).

Answer 1

The Correct Option is C

Step 1: Evaluate the expression for $n=1$:
Consider the expression $E(n) = 2^{4n+3} + 3^{3n+1}$.
Substituting $n = 1$, \[ E(1) = 2^7 + 3^4 = 128 + 81 = 209 \] Factorizing $209$, we get \[ 209 = 11 \times 19 \] Thus, the possible common divisors are $11$ and $19$.

Step 2: Evaluate the expression for $n=2$:
Now substitute $n = 2$, \[ E(2) = 2^{11} + 3^7 = 2048 + 2187 = 4235 \] Check divisibility:
For 11: Using the alternating sum rule, \[ 5 - 3 + 2 - 4 = 0 \] So, $4235$ is divisible by $11$.
For 19: \[ 4235 \div 19 \neq \text{integer} \] Hence, $4235$ is not divisible by $19$.
Therefore, the only common divisor is $11$.

Step 3: Nature of the divisor:
The number $11$ is both odd and prime.
Thus, $P$ is an odd prime number.
Hence, the correct option is (C).