Find the smallest number by which each of the following numbers must be to obtain a perfect cube. (1) 243 (1) 256 (1) 72 (v) 100 (iv) 67
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Find the smallest number by which each of the following numbers must be to obtain a perfect cube. (1) 243 (1) 256 (1) 72 (v) 100 (iv) 67
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Q2) Find the smallest number by which each of the following numbers must be multiplied to obtain a perfect cube:
(i)243
(ii)256
(iii)72
(iv)675
(v)100
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Solution:
(i) 243
Prime factors of 243 =
Here 3 do not appear in 3’s group.
Therefore, 243 must be multiplied by 3 to make it a perfect cube.
(ii) 256
Prime factors of 256 = 2\times2\times2\times2\times2\times2\times2\times22×2×2×2×2×2×2×2
Here one factor 2 is required to make a 3’s group.
Therefore, 256 must be multiplied by 2 to make it a perfect cube.
(iii) 72
Prime factors of 72 = 2\times2\times2\times3\times32×2×2×3×3
Here 3 does not appear in 3’s group.
Therefore, 72 must be multiplied by 3 to make it a perfect cube.
(iv) 675
Prime factors of 675 = 3\times3\times3\times5\times53×3×3×5×5
Here factor 5 does not appear in 3’s group.
Therefore 675 must be multiplied by 3 to make it
a perfect cube.
(v) 100
Prime factors of 100 = 2\times2\times5\times52×2×5×5
Here factor 2 and 5 both do not appear in 3’s group.
Therefore 100 must be multiplied by 2\times52×5= 10 to make it a perfect cube.