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If {x} is the product of all even integers from 1 to x inclusive, what

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Manager
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If {x} is the product of all even integers from 1 to x inclusive, what [#permalink]

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25 Oct 2016, 17:11
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If {x} is the product of all even integers from 1 to x inclusive, what is the greatest prime factor of {22} + {20}?

A) 23
B) 20
C) 11
D) 5
E) 2
[Reveal] Spoiler: OA

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Manager
Joined: 28 Jun 2016
Posts: 207
Concentration: Operations, Entrepreneurship
Re: If {x} is the product of all even integers from 1 to x inclusive, what [#permalink]

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25 Oct 2016, 17:24
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duahsolo wrote:
If {x} is the product of all even integers from 1 to x inclusive, what is the greatest prime factor of {22} + {20}?

A) 23
B) 20
C) 11
D) 5
E) 2

{22} + {20} = 22*{20} +{20} = 23*{20}

A
Manager
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Re: If {x} is the product of all even integers from 1 to x inclusive, what [#permalink]

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19 Aug 2017, 08:24
{22} + {20} = 22*{20*18*-------*2} + {20*18*16-------*2} = 23*{20}

Hence 23
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Re: If {x} is the product of all even integers from 1 to x inclusive, what [#permalink]

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19 Aug 2017, 09:18
duahsolo wrote:
If {x} is the product of all even integers from 1 to x inclusive, what is the greatest prime factor of {22} + {20}?

A) 23
B) 20
C) 11
D) 5
E) 2

hi..

two ways..

1) as given above..
{22} + {20} = 22*(20*18......*2)+(20*18....*2)=(22+1)(20*18...*2)=23{20}

2) {22} + {20}
$$2*4*.....*22+2*4...20 = 2^{11}*(1*2*3...*11)+2^{10}*(1*2*3....*10) = 2^{11}*11!+2^{10}*10! = 2*2^{10}*11*10!+2^{10}*10!=2^{10}*10!*(2*11+1)=2^{10}*10!*23$$
so 23
A
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Re: If {x} is the product of all even integers from 1 to x inclusive, what [#permalink]

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18 Oct 2017, 23:06
{X} is the product of all even integers including x this statement means we have to include in the set even numbers till x. ie if x =5 then even integers from 1 till 5 are 2 and 4 and {5} is the product of 2 and 4.
Now to the question.
{22}+{20}
This can be written as {2*4*6*8*10*12*14*16*18*20*22}+{2*4*6*8*10*12*14*16*18*20} (*=multiplication)
Now in both the curly bracket the common numbers are from 2 to 20. let us take them out.
2*4*6*8*10*12*14*16*18*20(22+1)=2*4*6*8*10*12*14*16*18*20(23)
Numbers 2 to 20 are divisible by 2 and 23 in the brackets is divisible by 23 the prime number given in the answer choices. Hence correct answer is C
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Re: If {x} is the product of all even integers from 1 to x inclusive, what [#permalink]

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24 Oct 2017, 06:07
duahsolo wrote:
If {x} is the product of all even integers from 1 to x inclusive, what is the greatest prime factor of {22} + {20}?

A) 23
B) 20
C) 11
D) 5
E) 2

{22} = 22 x 20 x 18 x 16 x … 4 x 2

{20} = 20 x 18 x 16 x … 4 x 2

{22} + {20} = (22 x 20 x 18 x 16 x … 4 x 2) + (20 x 18 x 16 x … 4 x 2)

From the two terms above, we pull out the common factor of (20 x 18 x 16 x … 4 x 2), yielding:

{22} + {20} = 20 x 18 x 16 x … 4 x 2 x (22 + 1)

{22} + {20} = 20 x 18 x 16 x … 4 x 2 x 23

We see that the greatest prime factor is 23.

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Re: If {x} is the product of all even integers from 1 to x inclusive, what   [#permalink] 24 Oct 2017, 06:07
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