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Bunuel
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Let’s find the prime factorization of 900.
900 = 2^2 * 3^2 * 5^2. The number of positive factors of 900 is (2 + 1)(2 + 1)(2 + 1) = 27.
Now, a positive factor of 900 that is a multiple of 5 must have at least one factor of 5.
So, the number of positive factors of 900 that are not multiples of 5 is (2 + 1)(2 + 1)(0 + 1) = 9.
Therefore, the number of positive factors of 900 that are multiples of 5 is 27 - 9 = 18.
Answer C
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Bunuel
What is the number of positive factors of 900 that are multiples of 5?

A. 27
B. 21
C. 18
D. 15
E. 9
 


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900 = 2^2 * 3^2 * 5^2
Possible factors are formed by a combination of individual factors of the primes:
For 2: 2^0, 2^1, 2^2
For 3: 3^0, 3^1, 3^2
For 5: 5^0, 5^1, 5^2

Since we need the factors that are multiple of 5, the possible options for 5 must be 5^1 and 5^2 only
Thus, total number of combinations = 3 * 3 * 2 = 18
Answer C
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Hello! I hope you are well anyone could explain a bit more how to solve these problems? or really go step by step.
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