Let the two numbers be represented based on their given ratio of 1:7. We can express these numbers as $k$ and $7k$, where $k$ is a common factor.
The Least Common Multiple (LCM) of two numbers $a$ and $b$ is the smallest positive integer that is divisible by both $a$ and $b$. For numbers $k$ and $7k$, their LCM is $7k$. This is because $7k$ is divisible by $k$ (result is 7) and by $7k$ (result is 1).
We are given that the LCM of the two numbers is 721. Using the relationship derived above:
$7k = 721$
To find the value of $k$, we divide both sides of the equation by 7:
$k = \frac{721}{7}$
$k = 103$
Now that we have the value of $k$, we can find the two numbers:
The question asks for the sum of these two numbers. We add the numbers found:
Sum = $103 + 721$
Sum = $824$
Thus, the sum of the two numbers is 824.
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?