A computer is programmed to multiply consecutive even integers 2*4*6*8

Question

A computer is programmed to multiply consecutive even integers 2*4*6*8*…*n until the product is divisible by 2013, What is the value of n?

A. 22
B. 38
C. 62
D. 122
E. 672

Shrey08 · Accepted Answer

\frac{(2*4*6*8*…*n)}{2013}

2013 can be factorized = 3*11*61

\frac{(2*4*6*8*…*n)}{3*11*61}

the value of n should be such that all the integers in the denominator are cancelled.

For n = 122 we will get 3, 11, 61 at least once in the numerator.

n=122 = 61*2

\frac{(2*4*6*8*…*22*...*122)}{3*11*61}

here we can get a 3 from 6, 11 from 22 and 61 from 122

Option D

Hunter2021 · Answer

This also makes logical sense as all the prime factors are odd but we are looking for an even number. 
Since the denominator has to include all the prime factors of the numerator to be a divisor the numerator will have to have at least a multiple of 3, 11, 61.
Then we can already eliminate A,B,C as they are too small. 
122 works as its a multiple of 61 and and the prime factors 3 and 11 are covered by 6 and 22 in the numerator as mentioned above by Shrey08.
In my opinion E would also work but its not necessary as we are looking for the minimum required number.

Correct me if I am wrong. 
Cheers, Hunter

samarpan.g28 · Answer

I agree, 122=61*2 hence we would select that as the smallest possible even number that fits out of all the options.

Aniket03 · Answer

How do you find factors easily for such a big number? e.g. I had difficulty finding 61 as a factor.

Adarsh_24 · Answer

2^(n/2) * (n/2)! Should be divisible by 2013

2013= 3*7*11*61

Since there is no 2 in 2013. N/2! Must have all these numbers. So n/2 must be at least 61
So n=122.

Sanskar115 · Answer

Hi  Aniket03 ,
it may look scary at the beginning, but it is straightforward. Just deduce all the number's Prime factors, which is very easy to find if you know all the divisibility rules. PFA, below if you don't know divisibility rules. Lmk if you have any other doubts.

byjus com/maths/divisibility-rules/

PS: I am not able to post URLs, putting space in between, if anyone is using remove the space
I hope this helps!

Adarsh_24 · Answer

I believe a number can have prime number factors up to square root of it. So try 7,11,13,... in case all else fails.

Akshaynandurkar · Answer

To solve this question, we need to factorize the number 2013
2013 = 3 * 671 (use sq.root method. here 671 is close to 625 which is square of 25. so check for the prime number up 23 to check whether 671 is prime number or not/ 2,3,5,7 are out of picture . check next 11. it divides 671.

2013 = 3 * 11 * 61

now for even numbers' products = 2.4.6.8.....n we know 3 and 11 could divide 6 and 22 easily for 61 to divide the product, we will have to go up to 122 (2*61)
hence answer is 122

bumpbot · Answer

Automated notice from GMAT Club BumpBot:

A member just gave Kudos to this thread, showing it’s still useful. I’ve bumped it to the top so more people can benefit. Feel free to add your own questions or solutions.

This post was generated automatically.

A computer is programmed to multiply consecutive even integers 246*8

Prep Toolkit

Top 5 GMAT Debrief Videos

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

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards

GMAT Problem Solving (PS) Questions