On factorising 675 into prime factors, we get:
\[675 = 3 \times 3 \times 3 \times 5 \times 5\]
On grouping the factors in triples of equal factors, we get:
\[675 = \left\{ 3 \times 3 \times 3 \right\} \times 5 \times 5\]
It is evident that the prime factors of 675 cannot be grouped into triples of equal factors such that no factor is left over. Therefore, 675 is a not perfect cube. However, if the number is multiplied by 5, the factors can be grouped into triples of equal factors and no factor will be left over.
Thus, 675 should be multiplied by 5 to make it a perfect cube.
Text Solution
Solution : By using prime factorization we could find:<br>(i)`675`<br>`675=3^3times5^2`<br>As we can see that `5` is not cubed.<br>Hence, `5` is the smallest number by which `675` should be multiplied to make it a perfect cube.<br><br>(ii)`1323`<br>`1323=3^3times7^2`<br>As we can see that `7` is not cubed.<br>Hence, `7` is the smallest number by which `1323` should be multiplied to make it a perfect cube.<br>(iii)`2580`<br>`2560=2^9times5`<br>As we can see that `5` is not cubed.<br>Hence, `5times5=25` is the smallest number by which `2560` should be multiplied 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:
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
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