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

answer has to be C.

total number of factors of 385 and 24 is 64. (by breaking down into primefactors and calculating factors)

Prime numbers in the multiplication are 2 3 5 7 11

So the answer is 5/64

IMO : C

What portion of the set of unique factors of the product of 24 and 385 are prime factors?


factors of 24=1,2,3,4,6,8,12,24 = 8 factors
factors of 385=1, 5, 7, 11, 35, 55, 77, 385 =8 factors

factors of 24*385 = 8*8=64,

prime factors among all factors

2,3,5,7,11 =5

so solution = prime factors/Total factors

=5/64

Factors for 24 = 1,2,3,4,6,8,12,24
Factors for 385 = 1,5,7,11,35,55,77,385

Total number of prime factors in (24*385)=5 (2,3,5,7,11)
Looking at the options it can only be option C = \(\frac{5}{64}\)

What portion of the set of unique factors of the product of \(24\) and \(385\) are prime factors?

In oder to reach the correct answer choice faster, we don't actually need to multiply those numbers. We will merely prime factorize them and find the overall number of factors:

\(24=2^3*3\)

\(385=5*7*11\)

The product of all prime factors will give us the product of the numbers themselves:

\(24*385=2^3*3*5*7*11\)

Next, we need to add \(1\) to the exponents of all prime factors and then multiply the results:

\((3+1)*(1+1)*(1+1)*(1+1)*(1+1)=64\)

So \(24*385\) has overall \(64\) distinct factors, including \(5\) prime factors \(2, 3, 5, 7,\) and \(11\). Thus the portion of \(64\) unique factors (ratio of prime to all) that are prime is \(5/64\).

Hence C

Rule: num.factor(x) is equal to the product of the "power+1" of its prime factors

prime factors of (24)=(2ˆ3•3)
prime factors of (385)=(7•11•5)
num.factors(24•385)=(2ˆ3•3•7•11•5)=(3+1)(1+1)(1+1)(1+1)(1+1)=64
no. of primes / total num factors = 5 {2,3,7,11,5} /64

Answer (C).

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 number of unique factors= 4*2*2*2*2=64
Prime factors= 5

Probability= 5/64

IMO C

https://gmatclub.com/forum/download/fil ... 0a1b19362c

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!

24 = 2^3 * 3
385 = 5 * 7 * 11

total factors = (2+1)(1+1)(1+1)(1+1)(1+1) = 64

Prime factors = 2, 3, 5, 7, 11

5/64. Answer is C.

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