Question

Find the smallest whole number by which of the following numbers must be divided to obtain a perfect cube. a) 1024 b) 648.​

Answer 1

Answer:

1. Use the prime factorisation method.
2. Now, check if there are three pairs for the same digit like 2 × 2 × 2.
3. If there are three pairs for each that means it is a perfect cube but if there are not complete pairs than add the digits for making it a perfect cube like if there is 2 × 2 only then add one more two.

I hope this might help you with ☺

Answer 2

Answer:

(i)81

Prime factors of 81 = 3\times3\times3\times33×3×3×3

Here one factor 3 is not grouped in triplets.

Therefore 81 must be divided by 3 to make it a perfect cube.

(ii) 128

Prime factors of 128 = 2\times2\times2\times2\times2\times22×2×2×2×2×2

Here one factor 2 does not appear in a 3’s group.

Therefore, 128 must be divided by 2 to make it a perfect cube.

(iii) 135

Prime factors of 135 = 3\times3\times3\times53×3×3×5

Here one factor 5 does not appear in a triplet.

Therefore, 135 must be divided by 5 to make it a perfect cube.

(iv) 192

Prime factors of 192 = 2\times2\times2\times2\times2\times32×2×2×2×2×3

Here one factor 3 does not appear in a triplet.

Therefore, 192 must be divided by 3 to make it a perfect cube.

(v) 704

Prime factors of 704 = 2\times2\times2\times2\times2\times2\times112×2×2×2×2×2×11

Here one factor 11 does not appear in a triplet.

Therefore, 704 must be divided by 11 to make it a perfect cube.

Step-by-step explanation:

I hope it will be helpful for you