Question

$ 2^{3n} - 7n - 1 $ is divisible by

• A. 64
• B. 36
• C. 49
• D. 25

Hint:

To solve modular problems, it's often useful to directly substitute values into the expression and check for divisibility, especially for small integers.

Answer 1

The Correct Option is C

To determine if the expression \( 2^{3n} - 7n - 1 \) is divisible by 49, we should check its divisibility for several values of \( n \).

1. When \( n = 0 \):
   \( 2^{3 \times 0} - 7 \times 0 - 1 = 1 - 0 - 1 = 0 \).
   Clearly, 0 is divisible by 49.
2. When \( n = 1 \):
   \( 2^{3 \times 1} - 7 \times 1 - 1 = 8 - 7 - 1 = 0 \).
   Again, 0 is divisible by 49.
3. When \( n = 2 \):
   \( 2^{3 \times 2} - 7 \times 2 - 1 = 64 - 14 - 1 = 49 \).
   49 is divisible by 49.

We observe that for these values, the expression results in numbers divisible by 49. The behavior suggests a pattern continues for all positive integers \( n \).
Hence, the correct answer is 49, validating that the expression \( 2^{3n} - 7n - 1 \) is divisible by 49.

Answer 2

We are tasked with determining the value of \( n \) such that the expression \( 2^{3n} - 7n - 1 \) is divisible by a specific number. 
To do this, let's first explore the options and substitute various values of \( n \) into the given expression to check divisibility. 
We will test each option by substituting the values of \( n \) and calculating \( 2^{3n} - 7n - 1 \). 
Step 1: Substitute different values of \( n \)
Start by testing the values \( n = 1, 2, 3, 4, \ldots \), and check when \( 2^{3n} - 7n - 1 \) becomes divisible by 49. 
Step 2: Check divisibility
For \( n = 3 \): \[ 2^{3(3)} - 7(3) - 1 = 2^9 - 21 - 1 = 512 - 21 - 1 = 490 \] Clearly, 490 is divisible by 49. 
Thus, the correct answer is \( 49 \).