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# Let f(n) = the number of distinct factors n has. For example, f(20) =

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Manager
Joined: 13 Sep 2016
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Let f(n) = the number of distinct factors n has. For example, f(20) = [#permalink]

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11 Oct 2016, 05:02
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68% (02:05) correct 32% (01:50) wrong based on 130 sessions

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Let f(n) = the number of distinct factors n has. For example, f(20) = 6, because 20 has six factors(1,2,4,5,10 and 20). Which of the following products is equal to 225?

a) f(10).f(100)
b) f(100).f(1000)
c) f(1000).f(10000)
d) f(100).f(10000)
e) f(10).f(1000)

[Reveal] Spoiler: OA

Kudos [?]: 92 [0], given: 320

Math Expert
Joined: 02 Aug 2009
Posts: 5352

Kudos [?]: 6134 [4], given: 121

Let f(n) = the number of distinct factors n has. For example, f(20) = [#permalink]

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11 Oct 2016, 06:03
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alanforde800Maximus wrote:
Let f(n) = the number of distinct factors n has. For example, f(20) = 6, because 20 has six factors(1,2,4,5,10 and 20). Which of the following products is equal to 225?

a) f(10).f(100)
b) f(100).f(1000)
c) f(1000).f(10000)
d) f(100).f(10000)
e) f(10).f(1000)

Hi,

You don't have to do any calculations but require to know the property of factors..

A SQUARE has odd number of factors...

So 225 is ODD..
Now check the choices..
Only D is multiple of two squares so the number of factors will be odd*odd=odd..

Rest all choices will be EVEN

100=2^2*5^2..
Number of factors= (2+1)(2+1)=3*3..
10000=2^4*5^4..
Number of factors=(4+1)(4+1)=25..

Ans D
_________________

Absolute modulus :http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html

Kudos [?]: 6134 [4], given: 121

Manager
Joined: 13 Sep 2016
Posts: 101

Kudos [?]: 92 [0], given: 320

Re: Let f(n) = the number of distinct factors n has. For example, f(20) = [#permalink]

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11 Oct 2016, 06:13
chetan2u wrote:
alanforde800Maximus wrote:
Let f(n) = the number of distinct factors n has. For example, f(20) = 6, because 20 has six factors(1,2,4,5,10 and 20). Which of the following products is equal to 225?

a) f(10).f(100)
b) f(100).f(1000)
c) f(1000).f(10000)
d) f(100).f(10000)
e) f(10).f(1000)

Hi,

You don't have to do any calculations but require to know the property of factors..

A SQUARE has odd number of factors...

So 225 being square of 15 will have ODD number of factors..
Now check the choices..
Only D is multiple of two squares so the number of factors will be odd*odd=odd..

Rest all choices will be EVEN

Ans D

Thanks for reply. I am still struggling to understand the logic behind this. Can you elaborate more on this?

Kudos [?]: 92 [0], given: 320

Math Expert
Joined: 02 Aug 2009
Posts: 5352

Kudos [?]: 6134 [1], given: 121

Re: Let f(n) = the number of distinct factors n has. For example, f(20) = [#permalink]

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11 Oct 2016, 06:22
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alanforde800Maximus wrote:
chetan2u wrote:
alanforde800Maximus wrote:
Let f(n) = the number of distinct factors n has. For example, f(20) = 6, because 20 has six factors(1,2,4,5,10 and 20). Which of the following products is equal to 225?

a) f(10).f(100)
b) f(100).f(1000)
c) f(1000).f(10000)
d) f(100).f(10000)
e) f(10).f(1000)

Hi,

You don't have to do any calculations but require to know the property of factors..

A SQUARE has odd number of factors...

So 225 being square of 15 will have ODD number of factors..
Now check the choices..
Only D is multiple of two squares so the number of factors will be odd*odd=odd..

Rest all choices will be EVEN

Ans D

Thanks for reply. I am still struggling to understand the logic behind this. Can you elaborate more on this?

Hi..

Each number can be written as sum of two factors..
Example 10..1*10;2*5;.
But a square has one set which has same integer, so only one number comes out of that set..that is why ODD.
Example 36...1*36;2*18;3*12:4*9;6*6.... here 6*6 gives only one integer 6
_________________

Absolute modulus :http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html

Kudos [?]: 6134 [1], given: 121

SVP
Joined: 11 Sep 2015
Posts: 1910

Kudos [?]: 2752 [2], given: 364

Re: Let f(n) = the number of distinct factors n has. For example, f(20) = [#permalink]

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11 Oct 2016, 07:31
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Top Contributor
I'm assuming that the periods in your post are meant to represent multiplication.
I have replaced the periods with "x"

alanforde800Maximus wrote:
Let f(n) = the number of distinct factors n has. For example, f(20) = 6, because 20 has six factors(1,2,4,5,10 and 20). Which of the following products is equal to 225?

a) f(10) x f(100)
b) f(100) x f(1000)
c) f(1000) x f(10000)
d) f(100) x f(10000)
e) f(10) x f(1000)

NICE RULE
If N = (p^a)(q^b)(r^c)..., where p, q, r,...(etc.) are prime numbers, then the total number of positive divisors of N is equal to (a+1)(b+1)(c+1)...

Example: 14000 = (2^4)(5^3)(7^1)
So, the number of positive divisors of 14000 = (4+1)(3+1)(1+1) = (5)(4)(2) = 40

----------------------------------------

Let's test a few values:
100 = (2^2)(5^2)
So, the number of positive divisors of 100 = (2+1)(2+1) = (3)(3) = 9
So, f(100) = 9

1000 = (2^3)(5^3)
So, the number of positive divisors of 1000 = (3+1)(3+1) = (4)(4) = 16
So, f(1000) = 16

10000 = (2^4)(5^4)
So, the number of positive divisors of 10000 = (4+1)(4+1) = (5)(5) = 25
So, f(10000) = 25

Which of the following products is equal to 225?
225 = 9 x 25
= f(100) x f(10000)

[Reveal] Spoiler:
D

_________________

Brent Hanneson – Founder of gmatprepnow.com

Kudos [?]: 2752 [2], given: 364

Manager
Joined: 26 Jan 2016
Posts: 115

Kudos [?]: 24 [1], given: 55

Location: United States
GPA: 3.37
Re: Let f(n) = the number of distinct factors n has. For example, f(20) = [#permalink]

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11 Oct 2016, 08:08
1
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The easiest way to do this is to use the trick of breaking down everything into primes then looking at the exponents. Lets take 100 as an example

100 breaks down to 10*10 which is 2*5*2*5 this is 2²*5². The exponent trick is to break down a number into primes and then add one to each exponent, the multiply. So this would result in 3*3 or 9 factors.

If you do this for the numbers in the question you'll see 100 has 9 factors, 1000 has 16 factors, and and 100000 has 25 factors (these are all perfect squares).

So 9*25=225 hence D

Kudos [?]: 24 [1], given: 55

Manager
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Kudos [?]: 247 [1], given: 73

Re: Let f(n) = the number of distinct factors n has. For example, f(20) = [#permalink]

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12 Oct 2016, 01:08
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alanforde800Maximus wrote:
Let f(n) = the number of distinct factors n has. For example, f(20) = 6, because 20 has six factors(1,2,4,5,10 and 20). Which of the following products is equal to 225?

a) f(10).f(100)
b) f(100).f(1000)
c) f(1000).f(10000)
d) f(100).f(10000)
e) f(10).f(1000)

Any integer can be written as a product of prime numbers.

If $$N = p^{a} q^{b} r^{c}$$, then no. of factors of $$N = (a+1) \times (b+1) \times (c+1)$$.

$$10 = 2^{1} \times 5^{1}$$, No. of factors = (1+1)*(1+1) = 4
$$100 = 2^{2} \times 5^{2}$$, No. of factors = (2+1) * (2+1) = 9
$$1000 = 2^{3} \times 5^{3}$$, No. of factors = (3+1) * (3+1) = 16
$$10000 = 2^{4} \times 5^{4}$$, No. of factors = (4+1) * (4+1) = 25

Option (d) is the correct answer.

Kudos [?]: 247 [1], given: 73

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Re: Let f(n) = the number of distinct factors n has. For example, f(20) = [#permalink]

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16 Nov 2017, 06:47
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Re: Let f(n) = the number of distinct factors n has. For example, f(20) =   [#permalink] 16 Nov 2017, 06:47
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