Question

If the HCF of 210 and 55 is expressible in the form of $210\times 5+55y$, then find $y$.

Answer 1

Hint: We will apply Euclid's division algorithm to find HCF of 210 and 55 and then we will use equations formed in the algorithm to evaluate the equation of the form \[210\times 5+55y.\] To obtain HCF of 210 and 55, we will use Euclid's division algorithm which uses Euclid's division lemma to find HCF of two numbers.

Complete step-by-step solution
According to Euclid's division lemma, for any two positive integers $a$ and $b$, there exits unique integers $q$ and $r$ satisfying $a=bq+r, 0\le r< b.$
Now, to find HCF of two numbers $a$ and $b$ such that $a >b$, using Euclid's division algorithm, we write$a=bq+r, 0\le r< b$. If $r=0$, then $b$ is the HCF of $a$ and $b$. If $r\ne 0$ , then we apply Euclid's division lemma to $b$ and $r$. This process is continued until the remainder is zero, at that stage the divisor is the required HCF.
We will now apply this algorithm to 210 and 55 as follows:
Since, $210>55$, so we can write,
$210=55\times 3+45\cdots \cdots (i)$
Here, $45\ne 0$, so applying division lemma to 55 and 45, where 55>45, we get,
$55=45\times 1+10\cdots \cdots (ii)$
Again, $10\ne 0$, so applying division lemma to 45 and 10, where 45>10, we get,
$45=10\times 4+5\cdots \cdots \left( iii \right)$
Again, $5\ne 0$, so applying division lemma to 10 and 5, where 10>5, we get,
$10=5\times 2+0$
Since now the remainder is zero, therefore divisor at this stage, that is 5, is the HCF of 210 and 55.
Now, from equation $\left( iii \right)$, we have,
$45=10\times 4+5$
$\Rightarrow 5=45-10\times 4\cdots \cdots \left( iv \right)$
Also, from equation $(ii)$, we can write,
$10=55-45$
Using this value of 10 in equation $\left( iv \right)$, we get,
$5=45-(55-45)\times 4$
$\Rightarrow 5=45-55\times 4+45\times 4$
$\Rightarrow 5=45\times 5-55\times 4\cdots \cdots \left( v \right)$
Again, from equation $\left( i \right)$ we can write,
$45=210-55\times 3$
Using this value of 45 in equation $\left( v \right)$, we get,
$5=\left( 210-55\times 3 \right)\times 5-55\times 4$
$\Rightarrow 5=210\times 5-55\times 3\times 5-55\times 4$
Taking 55 common, we get,
$\Rightarrow 5=210\times 5+55\times \left( -3\times 5-4 \right)$
$\Rightarrow 5=210\times 5+55\times (-19)\cdots \cdots \left( vi \right)$
Comparing equation $\left( vi \right)$ with the given equation: $5=210\times 5+55y$, we get,
$y=-19$.
This is the required solution.

Note: In this question, you can also take another approach to find HCF (e.g. long division method) and after finding HCF, you can also directly put that value equal to the given equation as the equation is the H.C.F and solve it to find $y$. Here our aim was to convert the last expression of HCF in terms of the given expression so that we can get the value of y.