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

Question

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

Bunuel · Accepted Answer

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.

Jatin108 · Answer

can we expect such questions?

gasoline · Answer

Dear Jatin

Trust you're doing well.
I hope you understood the question and what it meant to teach you.
You're tested here on your understanding of functions, number pattern, exponents, and divisions.
If you're able to notice the pattern, the question becomes fairly quick to crack.
Such a question is possible but would be rare.
Thank you.

Kinshook · Answer

enigma123

Given: The sequence f(n) = (2n)! ÷ n! is defined for all positive integer values of n.

Asked: 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?

Maximum power of 2 in f(1) = 1
Maximum power of 2 in f(2) = 2
Maximum power of 2 in f(3) = 3
Maximum power of 2 in f(4) = 4
Maximum power of 2 in f(5) = 5
Maximum power of 2 in f(6) = 6
Maximum power of 2 in f(7) = 7
Maximum power of 2 in f(8) = 8
Maximum power of 2 in f(9) = 9
Maximum power of 2 in f(10) =10

Maximum power of 2 in f(1)*f(2)*....*f(10) = 1+2+3+...+10 = 10*11/2 = 55

The greatest factor of x = 2^55

​​​​​​​IMO E

findingmyself · Answer

In these questions, the game is to look at the answer and come back to the question. The approach is very short and simple.

KarishmaB · Answer

Look for the pattern.

f(1) = 2 (One 2 as factor)
f(2) = 3*4 (Two 2s as factors)
f(3) = 4*5*6 (Three 2s as factors)
f(4) = 5*6*7*8 (Four 2s as factors)
f(5) = 6*7*8*9*10 (Five 2s as factors)
and so on...

When we multiply them all, the 2s we will obtain will be  2^1*2^2*2^3*2^4*2^5*...*2^{10}=2^{1+2+3+...+10}=2^{55}

Answer (E)

DataGlass · Answer

I did mine this way but it took quite a bit of time.

We have,
f(n) = (2n)! ÷ n!
f(1) = 2
f(2) = 12 (2*6)
f(3) = 120 (2*6*10)
f(4) = 1680 (2*6*10*14)

Notice the pattern here which is just add 4 to the last number and multiply with the above. This way, for f(10), we will have:
f(10) = 2*6*10*14*18*22*26*30*34*38

Again, from the pattern itself, we can see that 
x = 2^10 * 6^9 * 10^8 * 14^7*........ 34^2 * 38^1

Finding the number of 2s in above numbers, we get a total of 55 2s.

Therefore, the greatest factor is 2^55.

thalbritter · Answer

f(n) = (2n)! / n!

Separate the even factors in (2n)!:
2 * 4 * 6 * ... * 2n = 2^n * n!

So:
(2n)! = (1 * 3 * 5 * ... * (2n - 1)) * 2^n * n!

Therefore:
f(n) = 2^n * (1 * 3 * 5 * ... * (2n - 1))

The remaining product is odd, so f(n) contains exactly n factors of 2.

For the first 10 terms, the total number of factors of 2 is:
1 + 2 + 3 + ... + 10 = 55

Thus, the greatest listed power of 2 that divides x is:
2^55

Answer: E.

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

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

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