The question asks for the least number that is exactly divisible by the first ten natural numbers. This is equivalent to finding the Least Common Multiple (LCM) of the numbers 1, 2, 3, 4, 5, 6, 7, 8, 9, and 10.
LCM = $2^3 \times 3^2 \times 5^1 \times 7^1$
LCM = $8 \times 9 \times 5 \times 7$
LCM = $72 \times 35$
LCM = $2520$
Therefore, the least number which is divisible by the first ten natural numbers is 2520.
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?