Write 540 as the product of its prime factors. p is the smallest possible integer such that 540p is a square number. Find √540p, giving your answer as the product of its prime fact... Write 540 as the product of its prime factors. p is the smallest possible integer such that 540p is a square number. Find √540p, giving your answer as the product of its prime factors.
Understand the Problem
The question involves factorizing the number 540 into its prime factors and finding the smallest integer p such that 540p is a perfect square. It further asks for the value of the square root of 540p presented as a product of its prime factors.
Answer
The prime factorization of \( \sqrt{540p} \) is \( 2^1 \times 3^2 \times 5^1 \).
Answer for screen readers
The prime factorization of ( \sqrt{540p} ) is:
$$ 2^1 \times 3^2 \times 5^1 $$
Steps to Solve
- Factorize 540 into prime factors
To factorize 540, we repeatedly divide by the smallest prime numbers.
[ 540 \div 2 = 270 ] [ 270 \div 2 = 135 ] [ 135 \div 3 = 45 ] [ 45 \div 3 = 15 ] [ 15 \div 3 = 5 ] [ 5 \div 5 = 1 ]
So the complete factorization of 540 is:
$$ 540 = 2^2 \times 3^3 \times 5^1 $$
- Determine the value of ( p ) for ( 540p ) to be a perfect square
For a number to be a perfect square, all prime factors must have even exponents.
- Prime factorization of 540:
- ( 2: 2 ) (even)
- ( 3: 3 ) (odd)
- ( 5: 1 ) (odd)
To make all the exponents even:
- We need one more factor of ( 3 ) (from ( 3^3 ) to ( 3^4 )).
- We need one more factor of ( 5 ) (from ( 5^1 ) to ( 5^2 )).
Thus,
$$ p = 3^1 \times 5^1 = 3 \times 5 = 15 $$
- Calculate ( \sqrt{540p} )
Now we express ( 540p ):
$$ 540p = 540 \times 15 = 2^2 \times 3^3 \times 5^1 \times 3^1 \times 5^1 = 2^2 \times 3^4 \times 5^2 $$
Now, take the square root:
$$ \sqrt{540p} = \sqrt{2^2 \times 3^4 \times 5^2} = 2^{2/2} \times 3^{4/2} \times 5^{2/2} = 2^1 \times 3^2 \times 5^1 $$
So we have:
$$ \sqrt{540p} = 2 \times 9 \times 5 $$
- Final expression of ( \sqrt{540p} )
Thus,
$$ \sqrt{540p} = 2^1 \times 3^2 \times 5^1 $$
The prime factorization of ( \sqrt{540p} ) is:
$$ 2^1 \times 3^2 \times 5^1 $$
More Information
The number 540 has interesting properties; it is the product of six consecutive integers, and it is also related to the concept of perfect squares. The smallest integer ( p ), which allows ( 540p ) to be a perfect square, relates directly to the odd exponents in its prime factorization.
Tips
- Misidentifying the prime factors. Always verify through repeated division.
- Forgetting to ensure all prime factors' exponents are even when determining ( p ).
- Not properly calculating the square root of a product of prime factors; remember to halve the exponents.
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