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Manager  S
Joined: 01 Oct 2018
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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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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
Manager  B
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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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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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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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1
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

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  [#permalink]

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2
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
Intern  B
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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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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
Intern  B
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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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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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1
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".

Manager  S
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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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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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3
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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1
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.
Manager  S
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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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24=2*2*2*3
385=5*7*11

total number of factors=7
unique factors=5
so correct option is A
Manager  G
Joined: 25 Jul 2018
Posts: 211
Re: What portion of the set of unique factors of the product of 24 and 385  [#permalink]

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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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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) $$\frac{7}{64}$$ This question was provided by Veritas Prep for the Game of Timers Competition 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  [#permalink]

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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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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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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! _________________
"You don't want to to look back and know you could've done better".
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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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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   [#permalink] 23 Aug 2019, 23:59

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