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If r = 2^3 * 5^2 * 7 and s = 2^2 * 3^2 * 5, which of the 18 May 2012, 17:36
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If r = 2^3 * 5^2 * 7 and s = 2^2 * 3^2 * 5, which of the following is equal to the greatest common divisor of r and s?

A. 2 * 5
B. 2^2 * 5
C. 2^3 * 5^2
D. 2*3*5*7
E. 2^3 * 3^2 *5^2 * 7
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B
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If r = 2^3 * 5^2 * 7 and s = 2^2 * 3^2 * 5, which of the 18 May 2012, 17:42
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GCD = product of prime factors raised to the least power =2^2 x 5

Option B
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If r = 2^3 * 5^2 * 7 and s = 2^2 * 3^2 * 5, which of the 18 May 2012, 17:45
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jsphcal
If r = 2^3 * 5^2 * 7 and s = 2^2 * 3^2 * 5, which of the following is equal to the greatest common divisor of r and s?

A. 2 * 5
B. 2^2 * 5
C. 2^3 * 5^2
D. 2*3*5*7
E. 2^3 * 3^2 *5^2 * 7
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To find the greatest common divisor of two positive integers multiply the common factors of these integers by picking the lowest power of the common factors.

So, the GCD of r=2^3*5^2*7 and s=2^2*3^2*5 is 2^2*5.

Answer: B.
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If r = 2^3 * 5^2 * 7 and s = 2^2 * 3^2 * 5, which of the 29 Nov 2016, 09:29
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jsphcal
If \(r = 2^3 * 5^2 * 7\) and \(s = 2^2 * 3^2 * 5\), which of the following is equal to the greatest common divisor of r and s?

A. 2 * 5
B. 2^2 * 5
C. 2^3 * 5^2
D. 2*3*5*7
E. 2^3 * 3^2 *5^2 * 7
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GCD of \(r = 2^3 * 5^2 * 7\) and \(s = 2^2 * 3^2 * 5\) \(= 2^2*5\)

Hence, answer will be straight , (B) \(2^2 * 5\)
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If r = 2^3 * 5^2 * 7 and s = 2^2 * 3^2 * 5, which of the 02 Dec 2016, 10:32
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jsphcal
If r = 2^3 * 5^2 * 7 and s = 2^2 * 3^2 * 5, which of the following is equal to the greatest common divisor of r and s?

A. 2 * 5
B. 2^2 * 5
C. 2^3 * 5^2
D. 2*3*5*7
E. 2^3 * 3^2 *5^2 * 7
Show more

To determine the greatest common divisor (or greatest common factor) of any two numbers, we multiply together common prime factors with the smaller exponent. We are given the following:

r = 2^3 * 5^2 * 7

AND

s = 2^2 * 3^2 * 5^1

As we can see, 2 and 5 are the common prime factors. For the prime factor 2, the smaller exponent is 2 and for 5, the smaller exponent is 1. Thus, the GCF of r and s is 2^2 x 5^1.

Answer: B
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If r = 2^3 * 5^2 * 7 and s = 2^2 * 3^2 * 5, which of the 25 Oct 2020, 01:58
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If \(r = 2^3 * 5^2 * 7\) and \(s = 2^2 * 3^2 * 5\)


=> GCD = \(2^2 * 5\) [Take the lowest power of all the numbers which are common in both]

Answer B
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If r = 2^3 * 5^2 * 7 and s = 2^2 * 3^2 * 5, which of the 25 Oct 2020, 03:40
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r = 2^3 * 5^2 * 7 and s = 2^2 * 3^2 * 5

GCD is the greatest of the various factors that are common to both r and s.

So, GCD = 2^2*5 = 20
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If r = 2^3 * 5^2 * 7 and s = 2^2 * 3^2 * 5, which of the 20 Apr 2022, 11:23
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For finding the Highest Common Factor (HCF) / Greatest Common Divisor (GCD), each number has to be prime factorized and then the lowest power of each prime factor is to be considered.

The question already provides us with the prime factorised form of the numbers, so that makes our job easier.

r = \(2^3\) * \(5^2\) * 7

s = \(2^2\) * \(3^2\) * 5

Observing both the numbers,
    Lowest power of 2 = 2 (present in s)
    Lowest power of 3 = 0, since there is no power of 3 in r
    Lowest power of 5 = 1 (present in s)
    Lowest power of 7 = 0, since there is no power of 7 in s

Therefore, GCD (r and s) = \(2^2\) * \(3^0\) * \(5^1\) * \(7^0\) = \(2^2\) * 5, since any non-zero value raised to the power of zero equals 1.

The correct answer option is B.
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If r = 2^3 * 5^2 * 7 and s = 2^2 * 3^2 * 5, which of the 09 Oct 2025, 14:12
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