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HCF of 136, 170 using Euclid's algorithm

HCF of 136, 170 by Euclid's Divison lemma method can be determined easily by using our free online HCF using Euclid's Divison Lemma Calculator and get the result in a fraction of seconds ie., 34 the largest factor that exactly divides the numbers with r=0.

Highest common factor (HCF) of 136, 170 is 34.

HCF(136, 170) = 34

Ex: 10, 15, 20 (or) 24, 48, 96,45 (or) 78902, 89765, 12345

HCF of

Determining HCF of Numbers 136,170 by Euclid's Division Lemma

Below detailed show work will make you learn how to find HCF of 136,170 using the Euclidean division algorithm. So, follow the step by step explanation & check the answer for HCF(136,170).

Here 170 is greater than 136

Now, consider the largest number as 'a' from the given number ie., 170 and 136 satisfy Euclid's division lemma statement a = bq + r where 0 ≤ r < b

Step 1: Since 170 > 136, we apply the division lemma to 170 and 136, to get

170 = 136 x 1 + 34

Step 2: Since the reminder 136 ≠ 0, we apply division lemma to 34 and 136, to get

136 = 34 x 4 + 0

The remainder has now become zero, so our procedure stops. Since the divisor at this stage is 34, the HCF of 136 and 170 is 34

Notice that 34 = HCF(136,34) = HCF(170,136) .

Therefore, HCF of 136,170 using Euclid's division lemma is 34.

FAQs on HCF of 136, 170 using Euclid's Division Lemma Algorithm

1. What is the HCF(136, 170)?

The Highest common factor of 136, 170 is 34 the largest common factor that exactly divides two or more numbers with remainder 0.


2. How do you find HCF of 136, 170 using the Euclidean division algorithm?

According to the Euclidean division algorithm, if we have two integers say a, b ie., 136, 170 the largest number should satisfy Euclid's statement a = bq + r where 0 ≤ r < b and get the highest common factor of 136, 170 as 34.


3. Where can I get a detailed solution for finding the HCF(136, 170) by Euclid's division lemma method?

You can get a detailed solution for finding the HCF(136, 170) by Euclid's division lemma method on our page.