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The smallest natural number by which 135 must be divided to obtain a perfect cube is
- a)5
- b)3
- c)15
- d)9
Correct answer is option 'A'. Can you explain this answer?
Verified Answer
The smallest natural number by which 135 must be divided to obtain a p...
we have 135 = 3 x 3 x 3 x 5
Grouping the prime factors of 135 into triples, we are left over with 5.
∴ 135 is not a perfect cube
Now, [135]divided by5 = [ 3 x 3 x 3 x 5] divided by5
or 27 = 3 x 3 x 3
i.e. 27 is a perfect cube.
Thus, the required smallest number is 5
Grouping the prime factors of 135 into triples, we are left over with 5.
∴ 135 is not a perfect cube
Now, [135]divided by5 = [ 3 x 3 x 3 x 5] divided by5
or 27 = 3 x 3 x 3
i.e. 27 is a perfect cube.
Thus, the required smallest number is 5
Most Upvoted Answer
The smallest natural number by which 135 must be divided to obtain a p...
Certainly! Let's break down the problem of finding the smallest natural number by which 135 must be divided to obtain a perfect cube.
Understanding Perfect Cubes
A perfect cube is a number that can be expressed as \( n^3 \), where \( n \) is an integer. To find the smallest divisor to make 135 a perfect cube, we need to analyze its prime factorization.
Prime Factorization of 135
First, we factor 135:
- \( 135 = 3^3 \times 5^1 \)
Analyzing Exponents
For a number to be a perfect cube, all the exponents in its prime factorization must be multiples of 3:
- In \( 3^3 \), the exponent 3 is already a multiple of 3.
- In \( 5^1 \), the exponent 1 is not a multiple of 3.
To make the exponent of 5 a multiple of 3, we need to increase it from 1 to 3.
Calculating the Required Divisor
To achieve this, we need to remove \( 5^{1} \) from the factorization. Since we want to make the exponent of \( 5 \) reach 0 (which is also a multiple of 3), we need to divide by \( 5^{1} \).
Thus, the smallest natural number by which 135 must be divided is:
- \( 5^{1} = 5 \)
Conclusion
So, the correct answer is option (A) 5. By dividing 135 by 5, we reach a perfect cube, \( \frac{135}{5} = 27 \), which is \( 3^3 \).
Understanding Perfect Cubes
A perfect cube is a number that can be expressed as \( n^3 \), where \( n \) is an integer. To find the smallest divisor to make 135 a perfect cube, we need to analyze its prime factorization.
Prime Factorization of 135
First, we factor 135:
- \( 135 = 3^3 \times 5^1 \)
Analyzing Exponents
For a number to be a perfect cube, all the exponents in its prime factorization must be multiples of 3:
- In \( 3^3 \), the exponent 3 is already a multiple of 3.
- In \( 5^1 \), the exponent 1 is not a multiple of 3.
To make the exponent of 5 a multiple of 3, we need to increase it from 1 to 3.
Calculating the Required Divisor
To achieve this, we need to remove \( 5^{1} \) from the factorization. Since we want to make the exponent of \( 5 \) reach 0 (which is also a multiple of 3), we need to divide by \( 5^{1} \).
Thus, the smallest natural number by which 135 must be divided is:
- \( 5^{1} = 5 \)
Conclusion
So, the correct answer is option (A) 5. By dividing 135 by 5, we reach a perfect cube, \( \frac{135}{5} = 27 \), which is \( 3^3 \).
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