Use Euclid’s algorithm to find the HCF of 4052 and 12576
Solution:
As we can be see that 12576 > 4052, therefore let 4052 be d and 12567 be c such that c = dq + r where r is remainder, 0 ≤ r < d
Step 1: Let’s apply the division lemma to 12576 and 4052.
12576 = 4052 × 3 + 420
Step 2: As we can see r is 420 ≠ 0, we will apply division lemma to 4052 and 420.
4052 = 420 × 9 + 272
Step 3: Consider the new divisor 420 and the new remainder 272. We will apply the division lemma.
420 = 272 × 1 + 148
Step 4: Consider the new divisor 272 and the new remainder 148.We will apply the division lemma.
272 = 148 × 1 + 124
Step 5: Consider the new divisor 148 and the new remainder 124.We will apply the division lemma.
148 = 124 × 1 + 24
Step 6: Consider the new divisor 124 and the new remainder 24. We will apply the division lemma.
124 = 24 × 5 + 4
Step 7: Consider the new divisor 24 and the new remainder 4.We will apply the division lemma.
24 = 4 × 6 + 0
Now, the remainder is zero and the divisor is 4, thus the HCF of 12576 and 4052 is 4.
This implies 4 = HCF (24, 4) = HCF (124, 24) = HCF (148, 124) = HCF (272, 148) = HCF (420, 272) = HCF (4052, 420) = HCF (12576, 4052)
☛ Check: NCERT Solutions for Class 10 Maths Chapter 1
Use Euclid’s algorithm to find the HCF of 4052 and 12576
Summary:
Using Euclid’s algorithm the HCF of 4052 and 12576 is 4
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