using the common factor method find hcf . 45, 75and 135
Answers
Answer:
(i) Factor of 45 = F_{45}=3\times3\times5F
45
=3×3×5
Factor of 75 = F_{75}=\ 3\times5\times5F
75
= 3×5×5
And Factor of 135 =F_{135}=3\times3\times3\times5F
135
=3×3×3×5
Now the common factors of 45, 75 and 135 = 3 and 5
H.C.F = 3\times53×5 = 15
(ii) Factor of 48 =F_{48}=2\times2\times2\times2\times3F
48
=2×2×2×2×3
Factor of 36 = F_{36}=2\times2\times3\times3F
36
=2×2×3×3
And factor of 96 = F_{96}=2\times2\times2\times2\times2\times3F
96
=2×2×2×2×2×3
Now the common factor of 48, 36 and 96 = 2, 2 and 3
H.C.F = 2\times2\times32×2×3 = 12
(iii) Factor of 66 =F_{66}=2\times3\times11F
66
=2×3×11
Factor of 33 = F_{33}=3\times11F
33
=3×11
Factor of 132 = F_{132}=2\times2\times3\times11F
132
=2×2×3×11
Now the common factor of 66, 33 and 132 = 3 and 11
H.C.F = 3\times113×11 = 33
(iv) Factor of 24 =2\times2\times2\times22×2×2×2
Factor of 36 = 2\times2\times72×2×7
Factor of 60 = F_{60}=2\times2\times3\times5F
60
=2×2×3×5
And factor of 132 = F_{132}=2\times2\times3\times11F
132
=2×2×3×11
Now the common factors of 24, 36, 60 and 132 = 2,2 and3
H.C.F = 2\times2\times3=122×2×3=12
(v) Factor of 30 = F_{30}=2\times3\times5F
30
=2×3×5
Factor of 60 = F_{60}=2\times2\times3\times5F
60
=2×2×3×5
Factor of 90 = F_{90}=2\times3\times3\times5F
90
=2×3×3×5
And factor of 105 = F_{105}=3\times5\times7F
105
=3×5×7
Now common factor of 30, 60, 90 and 105 = 3 and 5
H.C.F = 3\times53×5 = 15
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Answer:
Hcf of 45,75,135= 15
Step-by-step explanation:
Factors of 45: 1, 3, 5, 9, 15, 45
Factors of 75: 1, 3, 5, 15, 25, 75
Factors of 135: 1, 3, 5, 9, 15, 27, 45, 135
Therefore the greatest common digit in all is 15 so the HCF of 45,75 and 135 is 15.
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