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Manager  B
Joined: 13 Sep 2016
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GMAT 1: 800 Q51 V51 Let f(n) = the number of distinct factors n has. For example, f(20) =  [#permalink]

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Question Stats: 71% (02:41) correct 29% (02:39) wrong based on 185 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)

Please assist with above problem.
Math Expert V
Joined: 02 Aug 2009
Posts: 7974
Let f(n) = the number of distinct factors n has. For example, f(20) =  [#permalink]

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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)

Please assist with above problem.

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..
answer would be F(100).F(10000)=9*25=225..

Ans D _________________
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Manager  B
Joined: 13 Sep 2016
Posts: 115
GMAT 1: 800 Q51 V51 Re: Let f(n) = the number of distinct factors n has. For example, f(20) =  [#permalink]

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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)

Please assist with above problem.

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?
Math Expert V
Joined: 02 Aug 2009
Posts: 7974
Re: Let f(n) = the number of distinct factors n has. For example, f(20) =  [#permalink]

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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)

Please assist with above problem.

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

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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"

Please go back and edit your original post.

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)

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

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

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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)

Please assist with above problem.

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.
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GMAT 1: 800 Q51 V49 GRE 1: Q170 V170 Re: Let f(n) = the number of distinct factors n has. For example, f(20) =  [#permalink]

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Hi All,

In this prompt, you can use "design" of this question, and the answer choices, to your advantage. We're told to find a product of two values that equals 225. In this prompt, there aren't that many ways to get to that product in this way though:

1 x 225
5 x 45
9 x 25
15 x 15

Looking at the answer choices, we know that each of those numbers has MORE than 1 factor, so (1x225) is out. We also know that none of those answers is the product of the same term twice, so (15x15) is out.

f(10) = 1,2,5,10 = 4 terms - which isn't an option, so we can eliminate Answers A and E. f(100) clearly has more than 5 factors, so we're looking for an answer that is (9x25).

From here, you just have to factor down two of the three numbers: 100, 1000, 10000 - you'll either have the exact math OR you'll have one of the correct terms and the remaining term will be the other one that you need. Thus, you'll know which answer is the match.

GMAT assassins aren't born, they're made,
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Re: Let f(n) = the number of distinct factors n has. For example, f(20) =  [#permalink]

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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)

Please assist with above problem.

10 - 4 Factors (So, Option A & E ruled out).
100 - 9 Factors
1000 - 16 Factors (Option C & B ruled out).

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

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