
Use Euclid’s division algorithm to find the HCF of 210 and 55.
Answer
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Hint: Euclid’s division algorithm is the process to the HCF of any two numbers. Where one should be greater or lesser to another. They can’t be equal. If they are equal then it’s a trivial case.
Complete step-by-step answer:
The given integers are 210 and 55. Observe that $210 > 55$.
On applying the Euclid division lemma to the division lemma to 210 and 55, we get, $210 = 55 \times 3 + 45$. Since, the remainder $45 \ne 0$.
So, we apply the division lemma to the divisor 55 and remainder 45 to get, $55 = 45 \times 1 + 10$.
Now, again we apply the division lemma to the divisor and remainder, $45 = 10 \times 4 + 5$. Since the remainder is not zero.
We keep on applying it. $10 = 5 \times 2 + 0$.
Observe that, we got our remainder as zero. So, the divisor at this stage or the remainder at the previous stage that is 5 is the HCF of 210 and 55.
Note: Here, HCF stands for Highest common factor. Two numbers can have many factors but in the highest common factor, we chose only that factor which is common in both of them and the highest among themselves.
Complete step-by-step answer:
The given integers are 210 and 55. Observe that $210 > 55$.
On applying the Euclid division lemma to the division lemma to 210 and 55, we get, $210 = 55 \times 3 + 45$. Since, the remainder $45 \ne 0$.
So, we apply the division lemma to the divisor 55 and remainder 45 to get, $55 = 45 \times 1 + 10$.
Now, again we apply the division lemma to the divisor and remainder, $45 = 10 \times 4 + 5$. Since the remainder is not zero.
We keep on applying it. $10 = 5 \times 2 + 0$.
Observe that, we got our remainder as zero. So, the divisor at this stage or the remainder at the previous stage that is 5 is the HCF of 210 and 55.
Note: Here, HCF stands for Highest common factor. Two numbers can have many factors but in the highest common factor, we chose only that factor which is common in both of them and the highest among themselves.
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