`2 ^(15) -1` is divisible by

Options

A. 4
B. 10
C. 2
D. 7

Answer

To determine whether \( 2^{15} - 1 \) is divisible by any of the given options (4, 10, 2, and 7), we can follow these steps:

### Step 1: Calculate \( 2^{15} - 1 \)

First, we need to compute \( 2^{15} \):
\[ 2^{15} = 32768 \]
Now, subtract 1:
\[ 2^{15} - 1 = 32768 - 1 = 32767 \]

### Step 2: Check divisibility by each option

Now we will check if \( 32767 \) is divisible by each of the options provided.

#### Check divisibility by 4:
To check if a number is divisible by 4, we can look at the last two digits.
\[ \text{Last two digits of } 32767 = 67 \]
Since \( 67 \div 4 = 16.75 \) (not an integer), \( 32767 \) is **not divisible by 4**.

#### Check divisibility by 10:
A number is divisible by 10 if it ends with 0.
\[ \text{Last digit of } 32767 = 7 \]
Since it does not end with 0, \( 32767 \) is **not divisible by 10**.

#### Check divisibility by 2:
A number is divisible by 2 if it is even.
\[ \text{Last digit of } 32767 = 7 \]
Since 7 is odd, \( 32767 \) is **not divisible by 2**.

#### Check divisibility by 7:
Now we will divide \( 32767 \) by 7:
\[ 32767 \div 7 \approx 4681 \]
Calculating \( 7 \times 4681 = 32767 \), we find that it is an exact division. Thus, \( 32767 \) is **divisible by 7**.

### Conclusion
The only option that \( 2^{15} - 1 \) is divisible by is **7**.

---