Question

Find H.C.F and L.C.M of the following numbers 8, 9 and 25.

Answer 1

Hint: To find the H.C.F and L.C.M of these numbers you have to find the prime factorization of 8, 9 and 25. Then the numbers which are common among all the three is H.C.F and L.C.M is the lowest common multiple which is divisible by all the given numbers. It is calculated by multiplying H.C.F by the uncommon factors of two numbers.

Complete step-by-step answer:
We have given the three numbers in the above problem of which we have to find the H.C.F and L.C.M:
8, 9 and 25
Prime factorizations of these three numbers are:
$\begin{array}{rl}& 8=2\times 2\times 2\times 1\\ & 9=3\times 3\times 1\\ & 25=5\times 5\times 1\end{array}$
As you can see the prime factorizations of the above three numbers you can see that only 1 is common in all the three numbers so H.C.F of these three numbers is 1.
L.C.M of these three numbers is calculated by multiplying H.C.F by the uncommon factors of all the three numbers.
H.C.F of the three numbers we have calculated above as 1 and in the below, we are going to show the multiplication of uncommon factors.
$2\times 2\times 2\times 3\times 3\times 5\times 5$
The multiplication of the above factors is equal to:
1800
Now, we are going to multiply 1800 by 1 to get the L.C.M of the three numbers.
$\begin{array}{rl}& 1800\left(1\right)\\ & 1800\end{array}$
Hence, the L.C.M of these three numbers is 1800.
Hence, L.C.M of 8, 9 and 25 is 1800 and H.C.F of 8, 9 and 25 is 1.

Note: Instead of finding the L.C.M in the way shown above, we can find the L.C.M of the three numbers by using the relation between L.C.M and H.C.F and multiplication of three numbers.
The relation between L.C.M and H.C.F and the multiplication of three numbers is:
$H.C.F\times L.C.M=8\times 9\times 25$
We have calculated the H.C.F as 1 so substituting this value in the above equation we get,
$\begin{array}{rl}& \left(1\right)\times L.C.M=8\times 9\times 25\\ & \Rightarrow L.C.M=1800\end{array}$
Hence, we have got the same L.C.M as we have solved above.