HCF of 1296, 5040 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., 144 the largest factor that exactly divides the numbers with r=0.
Highest common factor (HCF) of 1296, 5040 is 144.
HCF(1296, 5040) = 144
Ex: 10, 15, 20 (or) 24, 48, 96,45 (or) 78902, 89765, 12345
Below detailed show work will make you learn how to find HCF of 1296,5040 using the Euclidean division algorithm. So, follow the step by step explanation & check the answer for HCF(1296,5040).
Here 5040 is greater than 1296
Now, consider the largest number as 'a' from the given number ie., 5040 and 1296 satisfy Euclid's division lemma statement a = bq + r where 0 ≤ r < b
Step 1: Since 5040 > 1296, we apply the division lemma to 5040 and 1296, to get
5040 = 1296 x 3 + 1152
Step 2: Since the reminder 1296 ≠ 0, we apply division lemma to 1152 and 1296, to get
1296 = 1152 x 1 + 144
Step 3: We consider the new divisor 1152 and the new remainder 144, and apply the division lemma to get
1152 = 144 x 8 + 0
The remainder has now become zero, so our procedure stops. Since the divisor at this stage is 144, the HCF of 1296 and 5040 is 144
Notice that 144 = HCF(1152,144) = HCF(1296,1152) = HCF(5040,1296) .
Therefore, HCF of 1296,5040 using Euclid's division lemma is 144.
1. What is the HCF(1296, 5040)?
The Highest common factor of 1296, 5040 is 144 the largest common factor that exactly divides two or more numbers with remainder 0.
2. How do you find HCF of 1296, 5040 using the Euclidean division algorithm?
According to the Euclidean division algorithm, if we have two integers say a, b ie., 1296, 5040 the largest number should satisfy Euclid's statement a = bq + r where 0 ≤ r < b and get the highest common factor of 1296, 5040 as 144.
3. Where can I get a detailed solution for finding the HCF(1296, 5040) by Euclid's division lemma method?
You can get a detailed solution for finding the HCF(1296, 5040) by Euclid's division lemma method on our page.