Find the H. C.F for 513, 1134 and 1215

Solution:

In this question no method is specified to find the H.C.F so

Here we find the H.C.F by Euclid division lemma:-

a=bq+r

Here a= 1215 , b = 1134

1215>1134>513

1215=1134(1)+81

1134=81(14)+0

remainder becomes 0,the Hcf of 1134 and 1215 is 81.

Then find the Hcf of 81 and 513

513=81(6)+27

81=27(3)+0

remainder becomes 0, HCF of 513 & 81 is 27.

Hence, the HCF of 513,1134 and 1215 is 27.

By Prime factorization method

1215= 3×3×3×3×3×5= 3^5 ×5¹

1134= 3×3×3×3×2×7= 3⁴×2¹×7¹

513= 3×3×3×19= 3³× 19¹

H.C.F= 3³= 3×3×3= 27

Hence, the HCF of 513,1134 and 1215 is 27.

Hope this will help you

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