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

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Re: What portion of the set of unique factors of the product of 24 and 385  [#permalink]

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New post 12 Jul 2019, 02:55
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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  [#permalink]

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New post 12 Jul 2019, 03:54
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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  [#permalink]

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New post 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  [#permalink]

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New post 12 Jul 2019, 04:08
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IMO-C

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

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Re: What portion of the set of unique factors of the product of 24 and 385  [#permalink]

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New post 12 Jul 2019, 04:46
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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  [#permalink]

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New post 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  [#permalink]

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New post 12 Jul 2019, 05:12
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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  [#permalink]

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New post 12 Jul 2019, 05:45
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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  [#permalink]

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New post 12 Jul 2019, 05:59
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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  [#permalink]

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New post 12 Jul 2019, 06:37
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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  [#permalink]

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New post Updated on: 12 Jul 2019, 06:43
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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

Originally posted by RJ7X0DefiningMyX on 12 Jul 2019, 06:39.
Last edited by RJ7X0DefiningMyX on 12 Jul 2019, 06:43, edited 1 time in total.
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Re: What portion of the set of unique factors of the product of 24 and 385  [#permalink]

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New post 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  [#permalink]

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New post 12 Jul 2019, 07:33
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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  [#permalink]

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New post 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) \(\frac{7}{64}\)


 

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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

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Re: What portion of the set of unique factors of the product of 24 and 385  [#permalink]

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New post 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  [#permalink]

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New post 22 Jul 2019, 01:19
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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 Rajeet123

Sure, 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  [#permalink]

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New post 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 Rajeet123

Sure, 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  [#permalink]

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New post 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???
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Re: What portion of the set of unique factors of the product of 24 and 385   [#permalink] 23 Aug 2019, 23:59

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