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# lt;n> stands for the product of all even integers from 2

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lt;n> stands for the product of all even integers from 2 [#permalink]

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27 Oct 2005, 06:53
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<n> stands for the product of all even integers from 2 to n.
For example, <4> = 2 * 4, <6> = 2 * 4 * 6.

What is the greatest prime factor of <20>+<22>?
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27 Oct 2005, 07:09
Ans: 463

Here is the solution:
<20> + <22> = <20> (1 + 21*22) = <20>*463

As 463 is a prime, that is the greatest prime factor of <20>+<22>
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27 Oct 2005, 07:14
dogbert wrote:

Here is the solution:
<20> + <22> = <20> (1 + 21*22) = <20>*463

As 463 is a prime, that is the greatest prime factor of <20>+<22>

<22>= 2*4*6*...*22
<20>= 2*4*6*...*20
<20>+<22>= 2*4*6....*20*(1+22)= 2*4*6*...*23
23 is the greatest prime.
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27 Oct 2005, 08:05
I get 23 too...

<20>(1+22)=<20>(23)
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27 Oct 2005, 08:56
Sorry... I should not have included 21 in my calculations as it is only for even numbers.... 23 is the right answer...
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27 Oct 2005, 19:40
Well, this was question number 4 in my actual GMAT!!
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28 Oct 2005, 02:24
I agree with 23, but I thought I'd give a longer explanation than has been posted so far, in case anybody's trying to figure out the reasoning:

<20>+<22>
(2^10)(10!) + (2^11)(11!)
(2^10)(10!) + (2)(2^10)(10!)(11)
(2^10)(10!)[1 + 2(11)]
(2^10)(10!)(23)

The largest prime factors in each part of this equation are:
2^10: 2
10!: 7
23: 23

Thus, 23 is the greatest prime factor

Last edited by BumblebeeMan on 28 Oct 2005, 04:10, edited 1 time in total.
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28 Oct 2005, 03:43
rahulraao wrote:
Well, this was question number 4 in my actual GMAT!!

Well...this question was in my pocket for monthes...

Should have posted this question earlier...

Anyways,

Good job, everyone.

The OA is 23.
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28 Oct 2005, 22:48
Can someone break this problem down even more for a moron like me??? I am still not seeing it. Thanks!!
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29 Oct 2005, 09:06
BumblebeeMan wrote:
I agree with 23, but I thought I'd give a longer explanation than has been posted so far, in case anybody's trying to figure out the reasoning:

<20>+<22>
(2^10)(10!) + (2^11)(11!)
(2^10)(10!) + (2)(2^10)(10!)(11)
(2^10)(10!)[1 + 2(11)]
(2^10)(10!)(23)

The largest prime factors in each part of this equation are:
2^10: 2
10!: 7
23: 23

Thus, 23 is the greatest prime factor

Could you explain how you got (2^10)(10!)?
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30 Oct 2005, 06:32
TeHCM wrote:
BumblebeeMan wrote:
I agree with 23, but I thought I'd give a longer explanation than has been posted so far, in case anybody's trying to figure out the reasoning:

<20>+<22>
(2^10)(10!) + (2^11)(11!)
(2^10)(10!) + (2)(2^10)(10!)(11)
(2^10)(10!)[1 + 2(11)]
(2^10)(10!)(23)

The largest prime factors in each part of this equation are:
2^10: 2
10!: 7
23: 23

Thus, 23 is the greatest prime factor

Could you explain how you got (2^10)(10!)?

<20> = 2*4*6*8*10*12*14*16*18*20
<22> = 2*4*6*8*10*12*14*16*18*20*22

<20>+<22> = 2*4*6*8*10*12*14*16*18*20 + 2*4*6*8*10*12*14*16*18*20*22
= (2*4*6*8*10*12*14*16*18*20*22)(1+22)
=(2*4*6*8*10*12*14*16*18*20*22)(23)
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30 Oct 2005, 06:32
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