If x - 2 3 - 3 - 5 2 and y - 2 3 - 3 3 - then find HCF of x - y

Given that, [ x= 2^ 3 × 3× 5^ 2; y= 2^ 3 × 3^ 3 ] The Highest common factor is = 2^3× 3=24 ...

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Prime Factorisation

If x = 2 3...

Question

If $x = 2^{3} \times 3 \times 5^{2}$ and $y = 2^{3} \times 3^{3},$ then find $H C F$ of $x, y$

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If x = 2³ x 3 x 5², y = 2² x 3², then HCF (x, y) is :

x = 2^{3} \times 3 \times 5^{2}

y = 2^{2} \times 3^{3}

L C M (x, y)

2^{3} \times 3^{3} \times 7^{2}

b = 2^{2} \times 3^{2} \times 5^{3},

HCF of $(2^{3} \times 3^{2} \times 5), (2^{2} \times 3^{3} \times 5^{2}) a n d (2^{4} \times 3 \times 5^{3} \times 7)$ is

(a) 30 (b) 48 (c) 60 (d)105

\frac{x + 4}{3} = \frac{y - 1}{5} = \frac{z + 3}{4} and \frac{x + 1}{1} = \frac{y - 4}{1} = \frac{z - 5}{2}

\frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{- 3} and \frac{x + 3}{- 1} = \frac{y - 5}{8} = \frac{z - 1}{4}

\frac{5 - x}{- 2} = \frac{y + 3}{1} = \frac{1 - z}{3} and \frac{x}{3} = \frac{1 - y}{- 2} = \frac{z + 5}{- 1}

\frac{x - 2}{3} = \frac{y + 3}{- 2}, z = 5 and \frac{x + 1}{1} = \frac{2 y - 3}{3} = \frac{z - 5}{2}

\frac{x - 5}{1} = \frac{2 y + 6}{- 2} = \frac{z - 3}{1} and \frac{x - 2}{3} = \frac{y + 1}{4} = \frac{z - 6}{5}

\frac{- x + 2}{- 2} = \frac{y - 1}{7} = \frac{z + 3}{- 3} and \frac{x + 2}{- 1} = \frac{2 y - 8}{4} = \frac{z - 5}{4}

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