find the smallest number by which 100 must be multiplied to obtain a perfect cube
Answers
Answer:
Answer should be 10...
Step-by-step explanation:
10 ×10 ×10 = 1000
So this way answer should be 10...
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Step-by-step explanation:
In the prime factorization of a
perfect cube every prime factor occurs 3
times.
To determine
whether a number is a perfect cube or not proceed as follows:
1.Find the prime
factors of the given number.
2. Make Group of
3 equal prime factors.
3. 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.
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Solution:
(i) Prime factors of 243 = (3 × 3 × 3) ×
3 × 3
Here, two 3s 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 is a perfect cube.
Hence, the smallest
natural number by which 243 should be multiplied to make it a perfect cube is
3.
(ii) Prime factors of 256 = (2 × 2 × 2) ×( 2 × 2 × 2) ×
2 × 2
Here, two 2s are left
which are not in a triplet. To make 256 a cube, one more 2 is required. so, we
multiply 256 by 2 to make it a perfect cube
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 it a perfect cube is
2.
(iii) Prime factors of
72 = (2 × 2 × 2) × 3 × 3
Here, two 3s are left
which are not in a triplet. To make 72 a cube, one more 3 is required. so, we
multiply 72 by 3 to make it a perfect cube.
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 it a perfect cube is 3.
(iv) Prime factors of 675 = (3 × 3 × 3) ×
5 × 5
Here, two 5s are left
which are not in a triplet. To make 675 a cube, one more 5 is required. so, we
multiply 675 by 5 to make it a perfect cube.
675 × 5 = (3 × 3
× 3 )× (5 × 5 × 5 )= 3375 is a perfect cube.
Hence, the smallest
natural number by which 675 should be multiplied to make it a perfect cube is
5.
(v) Prime factors of 100
= (2 × 2) × (5 × 5)
Here, two 2s and two
5s are left which are not in a triplet. To make 100 a cube, we require one more
2 and one more 5. so, we multiply 100 by (2 x5 =10) to make it a perfect cube
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 it a perfect cube is 2
× 5 = 10.
(v) Prime factors of 100
= (2 × 2) × (5 × 5)
Here, two 2s and two
5s are left which are not in a triplet. To make 100 a cube, we require one more
2 and one more 5. so, we multiply 100 by (2 x5 =10) to make it a perfect cube
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 it a perfect cube is 2
× 5 = 10.
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Hope this will help you....