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15 Jan 2012, 15:28
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E
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
75%
(hard)
Question Stats:
64% (03:10) correct
36%
(03:07)
wrong
based on 741
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
(A) 2^20
(B) 2^30
(C) 2^45
(D) 2^52
(E) 2^55
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
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
15 Jan 2012, 15:53
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enigma123
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).
Answer: E.
General Discussion
Jatin108
Joined: 09 Apr 2015
Last visit: 28 Mar 2025
Posts: 26
Given Kudos: 40
Location: India
Schools: Wharton '26 Columbia CBS'26
GMAT 1: 690 Q46 V38
16 Jul 2023, 09:43
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can we expect such questions?
