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.
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Hope this will help you....
