Question

If two positive integers A and B can be ex-pressed as A = xy3 and B = xiy2z; x, y being prime numbers, the LCM (A, B) is

Answer 1

Step-by-step explanation:

When getting LCM using indices, we get the highest indices of the two and when getting HCF we get the lowest power of the two unknowns.

$a = x^{3} y {}^{2}$

$b = xy ^{3}$

LCM

Comparing indices of x and y in numbers a and b.

a: x's index is 3 whereas y's index is 2

b: x's index is 1 and y's index is 3.

Comparing the two: the highest index of x is 3 and the highest index of y is 3.

LCM = The highest indices of the unknowns (x and y)

$lcm = x {}^{3} y {}^{3}$

HCF = The lowest indices of x and y.

The lowest index of x is 1 and the lowest index of y is 2.

$hcf = xy {}^{2}$

$ab = x {}^{3}y {}^{2} xy {}^{3}$

$= x {}^{4} y {}^{5}$

$hcf \times lcm = x {}^{3} y {}^{3} (xy {}^{2} ) = x {}^{4} y {}^{5}$

Thus,

ab = LCM × HCF

HOPE THIS HELPS YOU

PLEASE MARK AS BRAINIEST

Answer 2

Given:

The two integers A and B are expressed as A = xy³ and B = xiy²z

x and y are the prime numbers

To Find:

The LCM of A and B

Solution:

As we know while getting the Lcm using the indices, we get the highest indices of the two and when getting HCF we get the lowest power of the two unknowns.

It is given that,

a = x³y²

b = xy³

LCM

Comparing the indices of x and y in the given 'a' and 'b'

a - x index is 3 and y index is 2 on the other hand,

b - x index is 1 and y index is 3

Now, we will compare the two highest indexes of x is 3 and y is 3.

LCM is the highest index of x and y

lcm = x³y³

And HCF is the lowest index of x and y. The lowest index of x is 1 and the lowest index of y is 2.

Now, it is given B= xy²

HCF = xy²

Putting a and b,

ab = x³y²xy³

    = $x^{4}y^{5}$

Now, HCF × LCM =  x³y³(xy²)

                     Solving the squares and cubes of x.

                             = $x^{4} y^{5}$

Thus, AB = LCM × HCF