`(2^(51)+2^(52)+2^(53)+2^(54)+2^(55))` is divisible by

Options

A. A) `23`
B. B) `58`
C. C) `124`
D. D) `127`

Answer

To solve the expression \(2^{51} + 2^{52} + 2^{53} + 2^{54} + 2^{55}\), we can factor out the smallest power of 2, which is \(2^{51}\).

### Step 1: Factor out \(2^{51}\)
\[ 2^{51} + 2^{52} + 2^{53} + 2^{54} + 2^{55} = 2^{51}(1 + 2 + 2^2 + 2^3 + 2^4) \]

### Step 2: Simplify the expression inside the parentheses
Now we simplify the expression inside the parentheses:
\[ 1 + 2 + 2^2 + 2^3 + 2^4 = 1 + 2 + 4 + 8 + 16 \]
Calculating this gives:
\[ 1 + 2 = 3 \\ 3 + 4 = 7 \\ 7 + 8 = 15 \\ 15 + 16 = 31 \]
Thus, we have:
\[ 1 + 2 + 2^2 + 2^3 + 2^4 = 31 \]

### Step 3: Substitute back into the factored expression
Now substituting back, we get:
\[ 2^{51} \cdot 31 \]

### Step 4: Determine divisibility
Now we need to check the divisibility of \(2^{51} \cdot 31\).

- \(2^{51}\) is clearly divisible by \(2\) and all higher powers of \(2\).
- \(31\) is a prime number.

### Conclusion
Thus, the expression \(2^{51} + 2^{52} + 2^{53} + 2^{54} + 2^{55}\) is divisible by \(31\) and \(2^{51}\).

To find the options provided, we can check the divisibility of \(2^{51} \cdot 31\) against the options given:
- Option A: 23 (not a factor)
- Option B: 58 (not a factor)
- Option C: 124 (which is \(4 \cdot 31\) and is a factor)
- Option D: 127 (not a factor)

The answer is that the expression is divisible by **124**.