Question

Express the hcf of 210 and 55 as 210x + 55y where x, y are integers in two different ways.

Answer 1

see the answer and write in your notes

Answer 2

Answer:

The HCF of $210$ and $55$ as $210x+55y$ in two different ways are $5=210x+55y$ and $1=42x+11y$, where $x,y$ are integers.

Step-by-step explanation:

Step 1 of 3

Finding HCF of $210$ and $55$.

Consider the given integers.

$210$, $55$

Clearly, $210 > 55$.

Using division algorithm to $210$ and $55$, we get

$210=55\times3 +45$ . . . . . (1)

This implies the remainder $45\neq 0$.

Again, apply the division algorithm to the divisor $55$ and remainder $45$, we get

$55=45\times1 +10$ . . . . . (2)

This implies the remainder $10\neq 0$.

So, apply the division algorithm to the divisor $45$ and remainder $10$, we get

$45=10\times4 +5$ . . . . . (3)

This implies the remainder $5\neq 0$.

So, apply the division algorithm to the divisor $10$ and remainder $5$, we get

$10=5\times2 +0$ . . . . . (4)

Thus, the remainder is $0$.

Therefore, the HCF of $210$ and $55$ is the last divisor, i.e.,

$5$ is the HCF of $210$ and $55$.

Step 2 of 3

Expressing the HCF of $210$ and $55$ as $210x+55y$.

Rewrite equation (3) as follows:

⇒ $5=45-10\times4$

⇒ $5=45-(55-45)\times 4$ (As $55-44=10$)

Using distributive property, we get

⇒ $5=45-55\times 4 +45\times 4$

⇒ $5=45\times 5-55 \times 4$ . . . . . (5)

From (1), we have

$210-55\times3 =45$

Substitute the value of $45$ in the equation (4) as follows:

⇒ $5=(210-55\times3)\times 5-55 \times 4$

Using distributive property, we get

⇒ $5=210\times 5-55\times15-55 \times 4$

⇒ $5=210 (5)-55(19)$

Let $x=5$ and $y=19$. Then,

$5=210x+55y$ . . . . . . (6)

where $x,y$ are integers.

Step 3 of 3

Expressing HCF in second way.

Divide the equation (6) by $5$, we get

$1=42x+11y$, where $x,y$ are integers.

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