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

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Bunuel
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A computer is programmed to multiply consecutive even integers 2*4*6*8 [#permalink] New post   04 Feb 2022, 03:15
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48% (02:12) correct 52% (02:16) wrong based on 159 sessions
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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
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Shrey08
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Re: A computer is programmed to multiply consecutive even integers 2*4*6*8 [#permalink] New post   06 Feb 2022, 00:42
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Bunuel wrote:
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


\(\frac{(2*4*6*8*…*n)}{2013}\)

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
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Hunter2021
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Re: A computer is programmed to multiply consecutive even integers 2*4*6*8 [#permalink] New post   18 Feb 2022, 04:41
Shrey08 wrote:
Bunuel wrote:
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


\(\frac{(2*4*6*8*…*n)}{2013}\)

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



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
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Re: A computer is programmed to multiply consecutive even integers 2*4*6*8 [#permalink] New post   16 Jan 2024, 09:58
Hunter2021 wrote:
Shrey08 wrote:
Bunuel wrote:
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


\(\frac{(2*4*6*8*…*n)}{2013}\)

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



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


I agree, 122=61*2 hence we would select that as the smallest possible even number that fits out of all the options.
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Re: A computer is programmed to multiply consecutive even integers 2*4*6*8 [#permalink]
16 Jan 2024, 09:58
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