Product of three natural numbers is 24000 and their HCF is 10. How many such triplets of numbers are there?

Options

A. 5
B. 4
C. 7
D. 6

Answer

To solve the problem step by step, we need to find the number of triplets of natural numbers \( n_1, n_2, n_3 \) such that:

1. The product \( n_1 \times n_2 \times n_3 = 24000 \)
2. The highest common factor (HCF) of \( n_1, n_2, n_3 \) is 10.

### Step 1: Express the numbers in terms of their HCF
Since the HCF of the three numbers is 10, we can express them as:
- \( n_1 = 10a \)
- \( n_2 = 10b \)
- \( n_3 = 10c \)

where \( a, b, c \) are co-prime numbers (their HCF is 1).

### Step 2: Substitute into the product equation
Now we substitute these expressions into the product equation:
\[ (10a) \times (10b) \times (10c) = 24000 \]
This simplifies to:
\[ 1000abc = 24000 \]

### Step 3: Solve for \( abc \)
Dividing both sides by 1000 gives:
\[ abc = \frac{24000}{1000} = 24 \]

### Step 4: Find the co-prime triplets
Now, we need to find the sets of co-prime numbers \( (a, b, c) \) such that their product is 24. The possible combinations of \( a, b, c \) must also satisfy the condition that their HCF is 1.

### Step 5: List the factor combinations of 24
The factors of 24 are:
1. \( 1, 1, 24 \)
2. \( 1, 2, 12 \)
3. \( 1, 3, 8 \)
4. \( 1, 4, 6 \)
5. \( 2, 3, 4 \)

### Step 6: Check the HCF of each combination
Now we check the HCF of each combination:
1. \( (1, 1, 24) \) → HCF = 1
2. \( (1, 2, 12) \) → HCF = 1
3. \( (1, 3, 8) \) → HCF = 1
4. \( (1, 4, 6) \) → HCF = 1
5. \( (2, 3, 4) \) → HCF = 1

All combinations have an HCF of 1.

### Step 7: Count the valid combinations
We found 5 valid combinations of \( (a, b, c) \) that satisfy both conditions:
1. \( (1, 1, 24) \)
2. \( (1, 2, 12) \)
3. \( (1, 3, 8) \)
4. \( (1, 4, 6) \)
5. \( (2, 3, 4) \)

### Conclusion
Thus, the number of such triplets of natural numbers \( (n_1, n_2, n_3) \) is **5**.

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