Three numbers are in the ratio 4 : 13 : 9, and their LCM is 6084. Their HCF is:
13
Let the HCF be \(k\), so the three numbers are \(4k,\ 13k,\ 9k\).
The parts 4, 13 and 9 are pairwise coprime, so the LCM of the numbers is \(k \times \text{LCM}(4, 13, 9)\).
Compute \(\text{LCM}(4, 13, 9) = 4 \times 13 \times 9 = 468\).
Given the LCM is 6084: \(468k = 6084\).
Solving: \(k = \frac{6084}{468} = 13\).
Hence, the HCF of the three numbers is 13.
Similar Questions
- The H.C.F. and the L.C.M. of two numbers are 5 and 495, respectively. If one of the numbers is 55, find the other one.
- The HCF of two numbers is 16 and their LCM is 240. If one number is 48, find the other number.
- The sum of two numbers is 132 and their L.C.M. is 630. What are the two numbers?
- The LCM of the numbers 12.8 and 0.004 is:
- Find the least number which is divisible by first ten natural numbers.
- Find the HCF of ($3^{45} - 1$) and ($3^{35} - 1$).
- The HCF and LCM of two numbers are in the ratio of 1: 30 and the difference between the HCF and LCM is 493. Find the product of LCM and HCF.
- HCF of $\frac{1}{3}, \frac{3}{4}, \frac{4}{5}$ and $\frac{5}{6}$ is:
- Which of the following is the greatest number that, when dividing 183, 127 and 211, leaves the same remainder each time?
- If the LCM of $20x^3y^2$ and $10x^4y^4$ is $20x^4y^4$, find the HCF.
Important Questions from LCM and HCF
Six bells begin to toll together and toll, respectively, at intervals of 3, 4, 6, 7, 8 and 12 seconds. After how many seconds, will they toll together again?
A and B are two prime numbers such that A > B and their LCM is 209. The value of A 2 - B is:
Find the least number which when divided by 12, 18, 24 and 30 leaves 4 as remainder in each case, but when divided by 7 leaves no remainder.
Calculate the HCF of \(\frac{12}{5}\) , \(\frac{14}{15}\) and \(\frac{16}{17}\) .
Three numbers are in the proportion of 3 : 8 : 15 and their LCM is 8280. What is their HCF?
