To solve this problem, we use the fundamental relationship between two numbers and their Highest Common Factor (H.C.F.) and Least Common Multiple (L.C.M.). The relationship is stated as:
Product of the two numbers = H.C.F. × L.C.M.
Let the two numbers be $Number1$ and $Number2$. We are given:
Using the formula:
$ \text{Number1} \times \text{Number2} = \text{H.C.F.} \times \text{L.C.M.} $
Substitute the given values into the equation:
$ 55 \times \text{Number2} = 5 \times 495 $
First, calculate the product of the H.C.F. and L.C.M.:
$ 5 \times 495 = 2475 $
Now, the equation becomes:
$ 55 \times \text{Number2} = 2475 $
To find $Number2$, divide the product by the known number (55):
$ \text{Number2} = \frac{2475}{55} $
Performing the division:
$ \text{Number2} = 45 $
Therefore, the other number is 45.
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?