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The sequence f(n) = (2n)! ÷ n! is defined for all positive

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The sequence f(n) = (2n)! ÷ n! is defined for all positive [#permalink]

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15 Jan 2012, 15:28
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Question Stats:

65% (02:13) correct 35% (02:29) wrong based on 254 sessions

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The sequence f(n) = (2n)! ÷ n! is defined for all positive integer values of n. If x is defined as the product of the first 10 ten terms of this sequence, which of the following is the greatest factor of x?

(A) 2^20
(B) 2^30
(C) 2^45
(D) 2^52
(E) 2^55

[Reveal] Spoiler:
The sequence f(n)=(2n)!/n! is defined for all positive integer values of n.
Guys - as the OA is not provided this is how I solved this. Can you please let me know if my approach is correct or not?

f(1) = 2
f(2) = $$2^2$$ * 3
f(3) = $$2^3$$ * 3 * 5
.
.
.
f(10) = 2^10

Product of 10 terms = 2^1 * 2^2 * ...........................2^10 --------------------------------(1)
The above can be simplified as the base is same i.e. 2

1+2+3+4+5+.........10= 2^55 and hence the answer is E
[Reveal] Spoiler: OA

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The sequence f(n) = (2n)! ÷ n! is defined for all positive [#permalink]

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15 Jan 2012, 15:53
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enigma123 wrote:
The sequence f(n) = (2n)! ÷ n! is defined for all positive integer values of n. If x is defined as the product of the first 10 ten terms of this sequence, which of the following is the greatest factor of x?
(A) 2^20
(B) 2^30
(C) 2^45
(D) 2^52
(E) 2^55

Given: $$f(n)=\frac{(2n)!}{n!}$$, so $$f(1)=\frac{(2)!}{1!}$$, $$f(2)=\frac{(4)!}{2!}$$, $$f(3)=\frac{(6)!}{3!}$$, ...

$$x=\frac{(2)!}{1!}*\frac{(4)!}{2!}*\frac{(6)!}{3!}*...*\frac{(20)!}{10!}$$.

Now, try to figure out the pattern in powers of 2 instead of brute force (reducing, factoring out 2's,...).

$$\frac{(2)!}{1!}=2^1$$;

$$\frac{(4)!}{2!}=2^2*3$$;

$$\frac{(6)!}{3!}=2^2*5*6=2^3*something$$;

$$\frac{(8)!}{4!}=2^4*3*5*7=2^4*something$$;

...

$$\frac{(20)!}{10!}=2^{10}*something$$

Basically the power of 2 goes up by 1 with each step.

So the greatest factor of x (greatest power of 2 which is a factor of x) will be $$2^1*2^2*2^1*...*2^{10}=2^{1+2+3+4+5+6+7+8+9+10}=2^{55}$$ (mean*# of terms=5.5*10).

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Re: The sequence f(n) = (2n)! ÷ n! is defined for all positive [#permalink]

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21 Dec 2017, 22:00
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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Re: The sequence f(n) = (2n)! ÷ n! is defined for all positive   [#permalink] 21 Dec 2017, 22:00
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