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\times2
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\times3
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\times5
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\times5
Here factor 2 and 5 both do not appear in 3’s group.
Therefore 100 must be multiplied by 2\times5= 10 to make it a perfect cube.
Solution:
A number is a perfect cube only when each factor in the prime factorization of the given number exists in triplets. Using this concept, the smallest number can be identified.
[i] 243
243 = 3 × 3 × 3 × 3 × 3
= 33 × 32
Here, one group of 3's is not existing as a triplet. To make it a triplet, we need to multiply by 3.
Thus, 243 × 3 = 3 × 3 × 3 × 3 × 3 × 3 = 729 is a perfect cube
Hence, the smallest natural number by which 243 should be multiplied to make a perfect cube is 3.
[ii]
256 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2
= 23 × 23 × 2 × 2
Here, one of the groups of 2’s is not a triplet. To make it a triplet, we need to multiply by 2.
Thus, 256 × 2 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 = 512 is a perfect cube
Hence, the smallest natural number by which 256 should be multiplied to make a perfect cube is 2.
[iii] 72
72 = 2 × 2 × 2 × 3 × 3
= 23 × 32
Here, the group of 3’s is not a triplet. To make it a triplet, we need to multiply by 3.
Thus, 72 × 3 = 2 × 2 × 2 × 3 × 3 × 3 = 216 is a perfect cube
Hence, the smallest natural number by which 72 should be multiplied to make a perfect cube is 3.
[iv] 675
675 = 5 × 5 × 3 × 3 × 3
= 52 × 33
Here, the group of 5’s is not a triplet. To make it a triplet, we need to multiply by 5.
Thus, 675 × 5 = 5 × 5 × 5 × 3 × 3 × 3 = 3375 is a perfect cube
Hence, the smallest natural number by which 675 should be multiplied to make a perfect cube is 5.
[v] 100
100 = 2 × 2 × 5 × 5
= 22 × 52
Here both the prime factors are not triplets. To make them triplets, we need to multiply by one 2 and one 5.
Thus, 100 × 2 × 5 = 2 × 2 × 2 × 5 × 5 × 5 = 1000 is a perfect cube
Hence, the smallest natural number by which 100 should be multiplied to make a perfect cube is 2 × 5 =10
☛ Check: NCERT Solutions for Class 8 Maths Chapter 7
Video Solution:
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
NCERT Solutions for Class 8 Maths Chapter 7 Exercise 7.1 Question 2
Summary:
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 are [i] 3, [ii] 2, [iii] 3, [iv] 5, and [v] 10
☛ Related Questions:
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