For every natural number a, b, c and x, if `(x^a+1)(x^b+1)(x^c+1)` is the factor of `S=1+x+x^2+x^3+...+x^111` what is the HCF of a, b and c

Options

A. 7
B. 3
C. 14
D. can't be determined

Answer

To solve the problem, we need to find the highest common factor (HCF) of the natural numbers \( a \), \( b \), and \( c \) given that \( (x^a + 1)(x^b + 1)(x^c + 1) \) is a factor of \( S = 1 + x + x^2 + x^3 + ... + x^{111} \).

### Step-by-Step Solution

1. **Identify the Series**:
The series \( S = 1 + x + x^2 + ... + x^{111} \) is a geometric series.

2. **Use the Formula for the Sum of a Geometric Series**:
The formula for the sum of a geometric series is:
\[ S_n = \frac{a(r^n - 1)}{r - 1} \]
where \( a \) is the first term, \( r \) is the common ratio, and \( n \) is the number of terms. Here, \( a = 1 \), \( r = x \), and \( n = 112 \) (since it includes terms from \( x^0 \) to \( x^{111} \)).
Therefore, we have:
\[ S = \frac{1(x^{112} - 1)}{x - 1} = \frac{x^{112} - 1}{x - 1} \]

3. **Factor \( x^{112} - 1 \)**:
The expression \( x^{112} - 1 \) can be factored using the difference of squares and other factorization techniques:
\[ x^{112} - 1 = (x^{56} - 1)(x^{56} + 1) \]
Continuing to factor \( x^{56} - 1 \):
\[ x^{56} - 1 = (x^{28} - 1)(x^{28} + 1) \]
And further:
\[ x^{28} - 1 = (x^{14} - 1)(x^{14} + 1) \]
Thus, we have:
\[ x^{112} - 1 = (x^{56} + 1)(x^{28} + 1)(x^{14} + 1)(x^{7} - 1)(x^{7} + 1) \]

4. **Identify Factors**:
From the factorization, we can see that the factors of \( S \) include \( x^{56} + 1 \), \( x^{28} + 1 \), and \( x^{14} + 1 \). Therefore, we can assign:
- \( a = 56 \)
- \( b = 28 \)
- \( c = 14 \)

5. **Find the HCF of \( a \), \( b \), and \( c \)**:
Now we need to find the HCF of \( 56 \), \( 28 \), and \( 14 \).
- The prime factorization of \( 56 \) is \( 2^3 \times 7 \).
- The prime factorization of \( 28 \) is \( 2^2 \times 7 \).
- The prime factorization of \( 14 \) is \( 2^1 \times 7 \).

The common factors are \( 2^1 \) and \( 7^1 \). Therefore, the HCF is:
\[ HCF = 2^1 \times 7^1 = 14 \]

### Final Answer:
The HCF of \( a \), \( b \), and \( c \) is \( 14 \).