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When prime-factorized, \(165 = 3*5*11\) and \(300 = 2^3 * 3 * 5^2\)

The common factors of the numbers 165 and 300 are 3 and 5.
Since we have been asked to find the distinct common factors,
they will be 1,3,5 and 3*5(15) which is 4(Option C)
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Bunuel
How many distinct factors do 165 and 300 have in common?

A. 2
B. 3
C. 4
D. 5
E. 8

Carefully looking at the question stem, what is really being asked is the no of distinct factors of the GCD of 165 and 300

Now, we can easily get the GCD (also known as HCF) of 165 and 300, i.e. 15, factorising which we get 3^1 * 5^1

Now, to get the total no. of factors, (1+1)*(1+1) = 4, and hence, answer choice (C)
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HCF of 165 and 300 is 15.
No of distinct factors of 15 are 4.

C is correct.
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Bunuel
How many distinct factors do 165 and 300 have in common?

A. 2
B. 3
C. 4
D. 5
E. 8


165 = 5 x 33 = 5 x 11 x 3

300 = 12 x 25 = 2^2 x 3 x 5^2

Thus, the factors in common are 1, 3, 5, 15. (Recall that 1 is a factor of every number, even though it doesn’t play a role in the prime factorization of a number since it is not prime.)

Answer: C
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Bunuel
How many distinct factors do 165 and 300 have in common?

A. 2
B. 3
C. 4
D. 5
E. 8

165 = 5 * 33 = 5 * 3 * 11. Since 300 doesn't contain an 11, but is a multiple of 3 and 5, GCF(300,165)=15. A trick here is \(\frac{165}{5} = 165 * \frac{2}{10} = \frac{330}{10} = 33\)).

Since the GCF is 15, any common factor of 165 and 300 will be a factor of 15. Therefore 1, 3, 5, 15 are the common factors. We have 4, so pick C.

However, if the GCF was bigger, let's say 96, then counting would be more exhausting and we might miss a factor. In this case, we can find the number of factors of 96 systematically this way:

Prime factorize \(96 = 32*3 = 2^5 * 3\). When constructing a factor of 96, we can choose from 0-5 multiples of two in the product. For the factor of three, we can choose either 0 threes or 1 multiple of three. Picking 0 multiples of both two and three means we construct the factor "1". Therefore there are \((5+1)*(1+1) = 6*2 = 12\) options, thus 96 has 12 factors in total. We can abbreviate this method with (a+1)(b+1)(c+1) ... where a, b, and c are the powers of each factor.
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