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04 Apr 2022, 08:45
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
75%
(hard)
Question Stats:
44% (01:30) correct
56%
(01:44)
wrong
based on 130
sessions
History
Date
Time
Result
Not Attempted Yet
How many factors of \(2^4*3^8*7^4\) are perfect squares?
(A) 12
(B) 22
(C) 45
(D) 16
(E) None of these
(A) 12
(B) 22
(C) 45
(D) 16
(E) None of these
05 Apr 2022, 09:59
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This is a question on the concept of Factors of a composite number.
A composite number is a positive integer greater than 1, which has more than 2 factors. Therefore, a composite number can always be broken down into its constituent prime factors by the process of prime factorization.
If N is a composite number, upon prime factorization, it can be written as
N = \(P_1^x * P_2^y * P_3^z*…. \)
Where \(P_1, P_2, P_3\) are the prime factors of N and x, y, z are positive integral powers of these prime factors.
For such a number N, the number of factors is given by the expression (x + 1) (y+1) (z+1)…….
However, in this question, we have to find out the factors of the number\( 2^4 * 3^8 * 7^4\) such that the factors are perfect squares.
The factors which are perfect squares will have the powers of the prime factors to be even numbers. For example,\( 2^2\) is a factor of the given number and is also a perfect square.
Now, the powers of 2 which are perfect squares are \(2^0, 2^2 and 2^4\).
Similarly, the powers of 3 which are perfect squares are \(3^0, 3^2, 3^4, 3^6 and 3^8\).
Lastly, the powers of 7 which are perfect squares are \(7^0, 7^2 and 7^4\).
Let’s consider \(2^0.\) With how many numbers in the other two rows can it be multiplied to give us different factors of the given number? Clearly, there are 5 options in terms of powers of 3 and 3 options in terms of powers of 7; in all, there are a total of 5 * 3 = 15 ways of combining\( 2^0\) with the powers of 3 and 7 to obtain factors of the given number.
Since there are 3 powers of 2 viz. \(2^0, 2^2 and 2^4\), and each of them have 15 ways of getting combined with the powers of 3 and 7, there are a total of 45 such combinations. Each combination represents a factor of the given number.
Therefore, the given number has 45 factors, each of which is a perfect square.
The correct answer option is C.
A composite number is a positive integer greater than 1, which has more than 2 factors. Therefore, a composite number can always be broken down into its constituent prime factors by the process of prime factorization.
If N is a composite number, upon prime factorization, it can be written as
N = \(P_1^x * P_2^y * P_3^z*…. \)
Where \(P_1, P_2, P_3\) are the prime factors of N and x, y, z are positive integral powers of these prime factors.
For such a number N, the number of factors is given by the expression (x + 1) (y+1) (z+1)…….
However, in this question, we have to find out the factors of the number\( 2^4 * 3^8 * 7^4\) such that the factors are perfect squares.
The factors which are perfect squares will have the powers of the prime factors to be even numbers. For example,\( 2^2\) is a factor of the given number and is also a perfect square.
Now, the powers of 2 which are perfect squares are \(2^0, 2^2 and 2^4\).
Similarly, the powers of 3 which are perfect squares are \(3^0, 3^2, 3^4, 3^6 and 3^8\).
Lastly, the powers of 7 which are perfect squares are \(7^0, 7^2 and 7^4\).
Let’s consider \(2^0.\) With how many numbers in the other two rows can it be multiplied to give us different factors of the given number? Clearly, there are 5 options in terms of powers of 3 and 3 options in terms of powers of 7; in all, there are a total of 5 * 3 = 15 ways of combining\( 2^0\) with the powers of 3 and 7 to obtain factors of the given number.
Since there are 3 powers of 2 viz. \(2^0, 2^2 and 2^4\), and each of them have 15 ways of getting combined with the powers of 3 and 7, there are a total of 45 such combinations. Each combination represents a factor of the given number.
Therefore, the given number has 45 factors, each of which is a perfect square.
The correct answer option is C.
General Discussion
15 Feb 2025, 09:56
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