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The function f(n) for an integer n is defined as the number of

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New post 27 May 2018, 10:50
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The function f(n) for an integer n is defined as the number of positive integers that are factors of n.
For example, f(6) = 4 because 6 has 4 factors: 1, 2, 3, 6. What is the value of f(f(24^2))?

A. 2
B. 4
C. 6
D. 8
E. 12

Source: Experts Global

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New post 28 May 2018, 06:47
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nigina93 wrote:
I would go with B and here is why.
First take care of f(24^2) or 576 which has 21 factors (before 24 it has 10 factors(1,2,3,4,6,8,9,12,16,18) + 1 (24) itself +10 factors after 24).
Now, we are left with f(21), which has 4 factors (1,3,7,21), asnwer is B.
I wonder whether there is easier way of realizing 21 factors of 576, because I do list them manually and took longer than 2 mins, Bunuel?


For any number, in order to find the total number of factors of any numbers, we need to do the following steps

1. Prime factorize number to the form \(a^x*b^y*c^z\) where a,b,c are the prime numbers.
2. Total number of prime factors are (x+1)(y+1)(z+1)

Let's try and use this to find the number of factors of 576 which is \(24^2\)

Prime-factorizing \((576)^2\), we get \((24^2)^2 = (2^3*3)^2 = 2^6*3^2\).
The number of factors of \((576)^2\) can be given as \((6+1)(2+1) = 7*3 = 21\)

Therefore, the value for \(f(f(24^2)) = f(21)\) is 4(Option B)
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New post 28 May 2018, 06:41
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I would go with B and here is why.
First take care of f(24^2) or 576 which has 21 factors (before 24 it has 10 factors(1,2,3,4,6,8,9,12,16,18) + 1 (24) itself +10 factors after 24).
Now, we are left with f(21), which has 4 factors (1,3,7,21), asnwer is B.
I wonder whether there is easier way of realizing 21 factors of 576, because I do list them manually and took longer than 2 mins, Bunuel?
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New post 28 May 2018, 07:09
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nigina93 wrote:
I would go with B and here is why.
First take care of f(24^2) or 576 which has 21 factors (before 24 it has 10 factors(1,2,3,4,6,8,9,12,16,18) + 1 (24) itself +10 factors after 24).
Now, we are left with f(21), which has 4 factors (1,3,7,21), asnwer is B.
I wonder whether there is easier way of realizing 21 factors of 576, because I do list them manually and took longer than 2 mins, Bunuel?


There IS an easier way.

Given \(f(f(24^2))\)
For simplicity's sake let \(f(24^2)\)=z
∴ we have f(z). :-)
Now lets solve z = \(f(24^2)\) --> \(f( (8*3)^2 )\) --> \(f((2^3*3)^2)\) -- > \(f(2^6*3^2)\)
To get the number of factors just add 1 to the powers and multiply (6+1) (2+1) = 7*3
∴ z=7*3
f(z)= (1+1)(1+1) = 2*2 = 4
∵ \((7^1)(3^1)\)

∴ Answer B
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New post 28 May 2018, 07:29
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Hey, guys. Thanks a lot, this way is much easier
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New post 18 May 2019, 02:16
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gmatjindi wrote:
Manually finding factors are pretty time concuming method.



gmatjindi - Unfortunately, thats the only way to solve this problem.

The important takeaway in this problem is

1. Prime factorize number to the form \(a^x∗b^y∗c^z\) where a,b,c are the prime numbers.
2. Total number of prime factors can be calculated by using \((x+1)(y+1)(z+1)\)
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New post 27 May 2018, 19:56
B

24*24 has 21 factors
21 has 4 factors
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Re: The function f(n) for an integer n is defined as the number of  [#permalink]

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New post 15 May 2019, 11:13
this question seems to be tough question
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Re: The function f(n) for an integer n is defined as the number of  [#permalink]

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New post 15 May 2019, 11:51
nigina93 wrote:
Hey, guys. Thanks a lot, this way is much easier



How the answer can be B? yesterday it was C.
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Re: The function f(n) for an integer n is defined as the number of  [#permalink]

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New post 18 May 2019, 01:00
Manually finding factors are pretty time concuming method.
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Re: The function f(n) for an integer n is defined as the number of  [#permalink]

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New post 20 May 2019, 14:13
Thank you very much for your detailed explanation!
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Re: The function f(n) for an integer n is defined as the number of   [#permalink] 20 May 2019, 14:13
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