The HCF of three natural numbers x, y and z is 13. If the sum of x, y and z is 117, then how many ordered triplets (x, y, z) exist?

Options

A. 38
B. 27
C. 54
D. 55

Answer

To solve the problem step by step, we will follow the reasoning outlined in the video transcript.

### Step-by-Step Solution:

1. **Understanding the Problem:**
We are given that the HCF (Highest Common Factor) of three natural numbers \(x\), \(y\), and \(z\) is 13. This means that \(x\), \(y\), and \(z\) can be expressed as multiples of 13.

2. **Expressing the Variables:**
Let:
\[ x = 13a, \quad y = 13b, \quad z = 13c \]
where \(a\), \(b\), and \(c\) are natural numbers.

3. **Setting Up the Equation:**
According to the problem, the sum of \(x\), \(y\), and \(z\) is 117:
\[ x + y + z = 117 \]
Substituting the expressions for \(x\), \(y\), and \(z\):
\[ 13a + 13b + 13c = 117 \]

4. **Simplifying the Equation:**
Factor out 13 from the left side:
\[ 13(a + b + c) = 117 \]
Dividing both sides by 13 gives:
\[ a + b + c = \frac{117}{13} = 9 \]

5. **Finding the Number of Solutions:**
We need to find the number of ordered triplets \((a, b, c)\) such that \(a + b + c = 9\) where \(a\), \(b\), and \(c\) are natural numbers. To do this, we can use the "stars and bars" theorem in combinatorics.

6. **Applying the Stars and Bars Theorem:**
The number of ways to distribute \(n\) indistinguishable objects (stars) into \(r\) distinguishable boxes (variables) is given by the formula:
\[ \binom{n + r - 1}{r - 1} \]
In our case, \(n = 9\) (the total we want) and \(r = 3\) (the three variables \(a\), \(b\), and \(c\)):
\[ \text{Number of solutions} = \binom{9 + 3 - 1}{3 - 1} = \binom{11}{2} \]

7. **Calculating the Combination:**
Now we calculate \(\binom{11}{2}\):
\[ \binom{11}{2} = \frac{11 \times 10}{2 \times 1} = \frac{110}{2} = 55 \]

8. **Conclusion:**
Therefore, the number of ordered triplets \((x, y, z)\) that satisfy the given conditions is **55**.