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Re: What portion of the set of unique factors of the product of 24 and 385
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12 Jul 2019, 02:55
What portion of the set of unique factors of the product of 24 and 385 are prime factors? 24=2*2*2*3 = 2^3*3 385 = 5*7*11 24*385 = 2^3*3*5*7*11 #factors = 4*2*2*2*2 = 64 #primes = 5 Answ 5/64 C



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Re: What portion of the set of unique factors of the product of 24 and 385
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12 Jul 2019, 03:54
In order to solve this question, we need to break down that is to prime factorize 24 and 385. 24 can be written as product of 2^3 and 3. Number of its factors can be counted by multiplying its exponent (adding one to it first). So, (3+1)*2=8 factors. Likewise, 385 can be written as product of 5*7*11. Since all of these integers are in power of 1, adding 1 gives 2*2*2=8. So we have 8*8=64, overall 64 factors, but some of them repeat and we need only unique factors which are 2,3,5,7,11 OR 5/64, which is C



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Re: What portion of the set of unique factors of the product of 24 and 385
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12 Jul 2019, 04:02
Solution: Prime factorization of 24 X 385 will be \(2^3\) X 5 X 7 X 11 Finding the Number of Factors of an Integer First make prime factorization of an integer n=\(a^p\)∗\(b^q\)∗\(c^r\), where a, b, and c are prime factors of n and p, q, and r are their powers. The number of factors of n will be expressed by the formula (p+1)(q+1)(r+1). NOTE: this will include 1 and n itself. Hence the number of unique factors of 24 X 385 will be (3+1) (1+1) (1+1) (1+) = 64 factors Number of prime numbers the product contains is 5 Hence the answer is \(\frac{5}{64}\) i.e C
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Re: What portion of the set of unique factors of the product of 24 and 385
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12 Jul 2019, 04:08
IMOC
24 X 385 = 2^3 * 3 * 5 * 7 * 11
# of factors= 4*2*2*2*2= 64
Prime Factors= 5 (2;3;5;7;11)
Req Fraction= 5/64
Posted from my mobile device



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Re: What portion of the set of unique factors of the product of 24 and 385
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12 Jul 2019, 04:46
24x385=\(2^3.3^1.5^1.7^1.11^1\)
Total number of factors = (3+1)(1+1)(1+1)(1+1)(1+1) = 4x2x2x2x2 = 64
Number of prime factors = 5 (2,3,5,7 and 11)
Hence, the answer is 5/64



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Re: What portion of the set of unique factors of the product of 24 and 385
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12 Jul 2019, 05:02
The Unique Factors of the product of 24 and 385 are = 5*7*11*2*2*2*3*1 so total factors are 8 and prime factors are 6 so prime factors are 6/8 portion of total unique factors



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Re: What portion of the set of unique factors of the product of 24 and 385
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12 Jul 2019, 05:12
Answer must be C because 2^3,3,5,7,11 are prime factors of 24, 385. So unique factors are 5. Overall number of factors are 4*2*2*2*2=64, 5/64 C



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Re: What portion of the set of unique factors of the product of 24 and 385
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12 Jul 2019, 05:45
What portion of the set of unique factors of the product of 24 and 385 are prime factors?
24*385 = (2^3)*(3)*(5)*(7)*(11)
Total factors = 4*2*2*2*2 = 64
total prime factors = 5 (2,3,5,7,11)
So, the answer will be "5/64".
ANSWER: C



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Re: What portion of the set of unique factors of the product of 24 and 385
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12 Jul 2019, 05:59
24*385
3*2^3*5*11*7 Total factors = (2+1)(2)(2)(2)(2) = 64
Prime numbers:2,3,5,11,7 Therefore 5/64



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Re: What portion of the set of unique factors of the product of 24 and 385
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12 Jul 2019, 06:37
24 = 2^3 * 3 385 = 5 * 11 * 7
Product = 2^3 * 3 * 5 * 7 * 11 Total factors = 4*2*2*2*2 =64 ( using total factors of p^a * q^b = (a+1)*(b+1) )
unique prime factors = 2,3,5,7,11 > total 5
so answer = 5/64
(C)



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What portion of the set of unique factors of the product of 24 and 385
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Updated on: 12 Jul 2019, 06:43
Good Q. So here's I took on this one:
product of 24 and 385 = 2^3 * 3^1 * 5^1 * 7^1 * 11^1
Total number of factors = (3+1)*(1+1)*(1+1)*(1+1)*(1+1) = 4*2*2*2*2 = 64
Number of prime factors = 5 (i.e. 2, 3, 5, 7 and 11)
Hence, portion of the set of unique factors of the product of 24 and 385 that are prime factors = 5/64, and that's (C)  the correct answer choice



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Re: What portion of the set of unique factors of the product of 24 and 385
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12 Jul 2019, 06:41
24=2*2*2*3 385=5*7*11
total number of factors=7 unique factors=5 so correct option is A



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Re: What portion of the set of unique factors of the product of 24 and 385
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12 Jul 2019, 07:33
What portion of the set of unique factors of the product of 24 and 385 are prime factors?
24=2^3*3 385=5*7*11
24*385=2^3*3*5*7*11 the number of Prime factors is 5 the number of factors is (3+1)*(1+1)*(1+1)*(1+1)*(1+1)*(1+1)=64 The answer is 5/64.
The answer choice is C



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Re: What portion of the set of unique factors of the product of 24 and 385
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12 Jul 2019, 20:10
Bunuel wrote: What portion of the set of unique factors of the product of 24 and 385 are prime factors? (A) \(\frac{5}{7}\) (B) \(\frac{7}{32}\) (C) \(\frac{5}{64}\) (D) \(\frac{6}{64}\) (E) \(\frac{7}{64}\)
24 = 2^3*3 385 = 5*7*11 Product of 24 & 385 = 2^3*3*5*7*11 Total Positive Factors = (3 + 1)*(1 + 1)*(1 + 1)*(1 + 1)*(1 + 1) = 64 Prime Factors = {2, 3, 5, 7, 11} = 5 factors Portion of prime factors = 5/64 IMO Option C Pls Hit Kudos if you like the solution



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Re: What portion of the set of unique factors of the product of 24 and 385
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22 Jul 2019, 00:19
firas92 wrote: \(24*385 = 2*2*2*3*5*7*11 = 2^3*3^1*5^1*7^1*11^1\)
So number of factors of the product \(24*385 = (3+1)*(1+1)*(1+1)*(1+1)*(1+1) = 64\)
Out of the 64 factors, 2,3,5,7,11 are the prime factors
So, the portion of the set of unique factors of the product of 24 and 385 are prime factors = \(\frac{5}{64}\) Might sound silly, but could you please explain how you got the number of factors as 64? Is that some formula?
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Re: What portion of the set of unique factors of the product of 24 and 385
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22 Jul 2019, 01:19
Rajeet123 wrote: firas92 wrote: \(24*385 = 2*2*2*3*5*7*11 = 2^3*3^1*5^1*7^1*11^1\)
So number of factors of the product \(24*385 = (3+1)*(1+1)*(1+1)*(1+1)*(1+1) = 64\)
Out of the 64 factors, 2,3,5,7,11 are the prime factors
So, the portion of the set of unique factors of the product of 24 and 385 are prime factors = \(\frac{5}{64}\) Might sound silly, but could you please explain how you got the number of factors as 64? Is that some formula? Hi Rajeet123Sure, remembering this as a formula will definitely be useful. If a number \(N\) can be expressed as \(p1^a*p2^b*p3^c\) (prime factorization), where \(p1, p2\) and \(p3\) are distinct prime numbers, then the number of factors of \(N\) is \((a+1)*(b+1)*(c+1)\) Consider the example of \(72\). \(72=2^3*3^2\) By our formula, the number of factors of \(72\) should be \((3+1)*(2+1) = 4*3 = 12\) If we need to analyze this by counting methods, in how many ways can we divide \(2^3*3^2\) such that there is no remainder? In the denominator of the fraction \(\frac{2^3*3^2}{Denominator}\), we obviously cannot have any term other than 2 or 3. How many powers can 2 and 3 take? \(2\) can take powers \(0, 1, 2\) or \(3\)  \(4\) options \(3\) can take powers \(0, 1\) or \(2\)  \(3\) options By fundamental principles of counting, we can just multiply these options to arrive at the number of factors. That is again, \(4*3=12\) So the factors are \(2^0*3^0\), \(2^0*3^1\), \(2^0*3^2\), \(2^1*3^0\), \(2^1*3^1\), \(2^1*3^2\), \(2^2*3^0\), \(2^2*3^1\), \(2^2*3^2\), \(2^3*3^0\), \(2^3*3^1\), \(2^3*3^2\) Hope this is clear!



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Re: What portion of the set of unique factors of the product of 24 and 385
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22 Jul 2019, 01:39
firas92 wrote: Rajeet123 wrote: firas92 wrote: \(24*385 = 2*2*2*3*5*7*11 = 2^3*3^1*5^1*7^1*11^1\)
So number of factors of the product \(24*385 = (3+1)*(1+1)*(1+1)*(1+1)*(1+1) = 64\)
Out of the 64 factors, 2,3,5,7,11 are the prime factors
So, the portion of the set of unique factors of the product of 24 and 385 are prime factors = \(\frac{5}{64}\) Might sound silly, but could you please explain how you got the number of factors as 64? Is that some formula? Hi Rajeet123Sure, remembering this as a formula will definitely be useful. If a number \(N\) can be expressed as \(p1^a*p2^b*p3^c\) (prime factorization), where \(p1, p2\) and \(p3\) are distinct prime numbers, then the number of factors of \(N\) is \((a+1)*(b+1)*(c+1)\) Consider the example of \(72\). \(72=2^3*3^2\) By our formula, the number of factors of \(72\) should be \((3+1)*(2+1) = 4*3 = 12\) If we need to analyze this by counting methods, in how many ways can we divide \(2^3*3^2\) such that there is no remainder? In the denominator of the fraction \(\frac{2^3*3^2}{Denominator}\), we obviously cannot have any term other than 2 or 3. How many powers can 2 and 3 take? \(2\) can take powers \(0, 1, 2\) or \(3\)  \(4\) options \(3\) can take powers \(0, 1\) or \(2\)  \(3\) options By fundamental principles of counting, we can just multiply these options to arrive at the number of factors. That is again, \(4*3=12\) So the factors are \(2^0*3^0\), \(2^0*3^1\), \(2^0*3^2\), \(2^1*3^0\), \(2^1*3^1\), \(2^1*3^2\), \(2^2*3^0\), \(2^2*3^1\), \(2^2*3^2\), \(2^3*3^0\), \(2^3*3^1\), \(2^3*3^2\) Hope this is clear! Absolutely! Thanks a tonne!
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Re: What portion of the set of unique factors of the product of 24 and 385
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23 Aug 2019, 23:59
Kinshook wrote: What portion of the set of unique factors of the product of 24 and 385 are prime factors?
\((A) \frac{5}{7}\)
\((B) \frac{7}{32}\)
\((C) \frac{5}{64}\)
\((D) \frac{6}{64}\)
\((E) \frac{7}{64}\)
\(24 = 2^3*3\) 385 = 5*7*11 Product of 24 and 385 = 24*385=\(2^3*3*5*7*11\)
# of factors of 24*385 = 4*2*2*2 = 64 # of prime factors = 5 (2,3,5,7 & 11)
Portion of prime factors to total unique factors = \(\frac{5}{64}\)
IMO C Can you please explain this part again???




Re: What portion of the set of unique factors of the product of 24 and 385
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