Answer:
:
Two numbers 1848 and 3058, where 3058 > 1848
3058 = 1848 × 1 + 1210
1848 = 1210 × 1 + 638 [Using Euclid's division algorithm to the given number 1848 and 3058]
1210 = 638 × 1 + 572
638 = 572 × 1 + 66
572 = 66 × 8 + 44
66 = 44 × 1 + 22
44 = 22 × 2 + 0
Therefore HCF of 1848 and 3058 is 22.
HCF (1848 and 3058) = 22
Let us find the HCF of the numbers 1331 and 22.
1331 = 22 × 60 + 11
22 = 11 × 2 + 0
HCF of 1331 and 22 is 11
HCF (22, 1331) = 11
Hence the HCF of the three given numbers 1848, 3058 and 1331 is 11.
HCF (1848, 3058, 1331) = 11