Question

Find the smallest number by which each of the following numbers must be multiplied to obtain a perfect cube:
$
  {\text{A}}{\text{. 243}} \\
  {\text{B}}{\text{. 3072}} \\
  {\text{C}}{\text{. 11979}} \\
  {\text{D}}{\text{. 19652}} \\
$

Answer 1

Hint – To find the smallest number to be multiplied we find the prime factors of the given number and make groups of 3 equal prime factors.

Complete Step-by-Step solution:
A. Given: 243
Prime factors of 243 = (3 × 3 × 3) × 3 × 3
Here, two 3’s are left which are not in a triplet. To make 243 a cube, one more 3 is required.
So, we multiply 243 by 3 to make it a perfect cube.

 243 × 3 = (3 × 3 × 3) × (3 × 3 × 3) = 729
729 = 9 × 9 × 9 = ${9^3}$
Therefore, 729 is a perfect cube.

Hence, the smallest number that makes 243 a perfect cube = 3

B. Given: 3072

Prime factors of 3072 = ${2^{10}}$× 3
                               = (${2^3}$× ${2^3}$×${2^3}$) × 2 × 3

Here, one 2 and one 3 are left which are not in a triplet. To make 3072 a cube, two more 2’s and two more 3’s are required.
So, we multiply 3072 by two 2’s and two 3’s to make it a perfect cube.

 3072 × 2 × 2 × 3 × 3 = (${2^3}$×${2^3}$×${2^3}$) × 2 × 2 × 2 × 3 × 3 × 3
                                          = 110592
110592 = 48 × 48 × 48 = ${48^3}$
 Therefore, 11059 is a perfect cube.

Hence, the smallest number that makes 3072 a perfect cube = ${2^2}$× ${3^2}$
                                                     = 4 × 9 = 36

C. Given: 11979
Prime factors of 11979 = (11 × 11 × 11) × 3 × 3
Here, two 3s are left which are not in a triplet. To make 11979 a cube, one more 3 is required.
So, we multiply 11979 by 3 to make it a perfect cube.

 11979 = (11 × 11 × 11) × 3 × 3 × 3 = 35937

35937 = 33 × 33 × 33 = ${33^3}$
Therefore, 35937 is a perfect cube.
Hence, the smallest number that makes 35937 a perfect cube = 3

D. Given: 19652

Prime factors of 19652 = (17 × 17 × 17) × 2 × 2

Here, two 2s are left which are not in a triplet. To make 19652 a cube, one more 2 is required.

So, we multiply 19652 by 2 to make it a perfect cube.

19652 = (17 × 17 × 17) × 2 × 2 × 2 = 39304

39304 = 34 × 34 × 34 = ${34^3}$
Therefore, 39304 is a perfect cube.
Hence, the smallest number that makes 39304 a perfect cube = 2

Note – The key in solving such types of problems is to prime factorize the given number. To find the cube root making a triplet of 3 equal prime factors is an important step. If a group contains only one or two equal prime factors then a given number is not a perfect cube otherwise it is a perfect cube.
Prime number is a number which only has 1 and itself as factors.