CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

Find the smallest number by which each of the following numbers must be multiplied to obtain a perfect cube.
(i) 243
(ii) 256

Open in App
Solution

(i) The prime factorisation of 243 = 3×3×3×3×3 . Here, two 3s are extra which are not in a triplet.
To make 243 a cube, one more 3 is required.
In that case, 243×3=3×3×3×3×3×3=729 is a perfect cube.
Therefore, the smallest natural number by which 243 should be multiplied to make it a perfect cube is 3.

(ii) The prime factorisation of 256=2×2×2×2×2×2×2×2
Here, two 2s are extra which are not in a triplet.
To make 256 a cube, one more 2 is required.
Then, we obtain 256 × 2 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 = 512, which is a perfect cube.
Therefore, the smallest natural number by which 256 should be multiplied to make it a perfect cube is 2.

flag
Suggest Corrections
thumbs-up
3
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Prime Factorisation_Tackle
MATHEMATICS
Watch in App
Join BYJU'S Learning Program
CrossIcon