Answered

Q. LCM of two numbers is 51 times of HCF. If one number is 27 find the another number.
Answer:
Given one number is 27
LCM= HCF x 51
We know that LCM= HCF x Remaining factors,
Hence , LCM= HCFx 3x17. (As 51= 3x17)
As 3 and 17 are two prime numbers,
we can say the two numbers are....
Either HCF x 3 and HCF x 17.......(1)
Or FCF x 1 and HCF x 51.............(2)
For case (1).....
FCF x 3= 27 (given one number is 27 )
Or, HCF= 9
In that case another number is = 9 x 17= 153.
For case(2)..........
FCF x 1 = 27 ( given one number is 27)
Or, FCF= 27
In that case another number is = 27 x 17= 1377.
So the required answer is 153 or 1377.​

Answer :

bsunitha041982

Answer:

To find the other number when the LCM of two numbers is 51 times their HCF, and one number is 27, follow these steps:

1. **Understand the Relation**: The LCM (Least Common Multiple) of two numbers \(a\) and \(b\) is given by:

\[ \text{LCM}(a, b) = \text{HCF}(a, b) \times \frac{a \times b}{\text{HCF}(a, b)} \]

Here, the LCM is \(51 \times \text{HCF}(a, b)\).

2. **Given Values**: Let one number be \(a = 27\), and the LCM of \(a\) and \(b\) is \(51 \times \text{HCF}(27, b)\).

3. **Prime Factorization of 51**: We know \(51 = 3 \times 17\).

4. **Possible Cases**:

- The other number \(b\) could be \(3 \times \text{HCF}(27, b)\) and \(17 \times \text{HCF}(27, b)\).

- Alternatively, \(b\) could be \(\text{HCF}(27, b)\) and \(51 \times \text{HCF}(27, b)\).

5. **Finding the HCF**:

- For the case \(3 \times \text{HCF}(27, b) = 27\), solve for \(\text{HCF}(27, b)\). Here, \(\text{HCF}(27, b) = 9\). Therefore, the other number \(b = 9 \times 17 = 153\).

6. **Check Second Case**:

- If \(\text{HCF}(27, b) = 27\), then the other number \(b = 27 \times 17 = 459\).

7. **Conclusion**:

- Based on the calculation, the other number \(b\) can be \(153\) or \(459\). In your answer, you mentioned \(1377\), but this seems to be an error. The correct calculation leads to \(153\) or \(459\) as the other number \(b\).

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