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: Test the expression for $n=1$ to find potential divisors.
Let the expression be $E(n) = 2^{4n+3} + 3^{3n+1}$.
For $n = 1$,
$E(1) = 2^{4(1)+3} + 3^{3(1)+1} = 2^7 + 3^4 = 128 + 81 = 209.$
The prime factors of $209$ are $11 \times 19$.
Hence, the possible values of $P$ are $11$ or $19$.
Step 2: Test the expression for $n=2$ to identify the common divisor.
For $n = 2$,
$E(2) = 2^{4(2)+3} + 3^{3(2)+1} = 2^{11} + 3^7 = 2048 + 2187 = 4235.$
Now, check the divisibility of $4235$ by $11$ and $19$.
Divisibility by 11: The alternating sum of digits is $5 - 3 + 2 - 4 = 0$,
which means $4235$ is divisible by $11$.
Divisibility by 19: $4235 \div 19 = 222$ with a remainder of $17$.
Hence, $4235$ is not divisible by $19$.
Since $P$ must divide the expression for all values of $n$, the common factor is $11$.
Therefore, $P = 11$.
Step 3: Analyze the properties of $P = 11$.
The number $11$ is an odd integer.
The number $11$ is also a prime number.
Hence, $P$ is an odd prime integer.
Therefore, the correct answer is (C).