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The product of the first eight positive even integers is closest to

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The product of the first eight positive even integers is closest to  [#permalink]

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New post Updated on: 26 Dec 2017, 09:37
4
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A
B
C
D
E

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  75% (hard)

Question Stats:

56% (02:03) correct 44% (01:57) wrong based on 248 sessions

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The product of the first eight positive even integers is closest to which of the following powers of 10?

(A) \(10^9\)
(B) \(10^8\)
(C) \(10^7\)
(D) \(10^6\)
(E) \(10^5\)

I can get the correct answer (c) by doing actual multiplication of 2,4,6,8,10,12,14 and 16. Is there any other way this can be solved?

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Originally posted by enigma123 on 25 Feb 2012, 17:01.
Last edited by Bunuel on 26 Dec 2017, 09:37, edited 2 times in total.
Edited.
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Re: The product of the first eight positive even integers is closest to  [#permalink]

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New post 25 Feb 2012, 17:16
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1
enigma123 wrote:
The product of the fir…st eight positive even integers is closest to which of the following powers of 10?
(A) \(10^9\)
(B) \(10^8\)
(C) \(10^7\)
(D) \(10^6\)
(E) \(10^5\)

I can get the correct answer (c) by doing actual multiplication of 2,4,6,8,10,12,14 and 16. Is there any other way this can be solved?


We should find the approximate value of 2*4*6*8*10*12*14*16 to some power of 10.

2*4*6=~50 (actually less than 50);
8*12=~100 (actually less than 100);
10;
14*16=~200 (actually more than 200);

2*4*6*8*10*12*14*16=~50*100*10*200=10^7 (6 zeros plus one zero produced by 2*5).

Answer: C.

Similar question to practice: the-product-of-all-the-prime-numbers-less-than-20-is-closest-126587.html

Hope it helps.
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Re: The product of the first eight positive even integers is closest to  [#permalink]

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New post 22 Nov 2013, 07:29
Bunuel wrote:
enigma123 wrote:
The product of the fir…st eight positive even integers is closest to which of the following powers of 10?
(A) \(10^9\)
(B) \(10^8\)
(C) \(10^7\)
(D) \(10^6\)
(E) \(10^5\)

I can get the correct answer (c) by doing actual multiplication of 2,4,6,8,10,12,14 and 16. Is there any other way this can be solved?


We should find the approximate value of 2*4*6*8*10*12*14*16 to some power of 10.

2*4*6=~50 (actually less than 50);
8*12=~100 (actually less than 100);
10;
14*16=~200 (actually more than 200);

2*4*6*8*10*12*14*16=~50*100*10*200=10^7 (6 zeros plus one zero produced by 2*5).

Answer: C.

Similar question to practice: the-product-of-all-the-prime-numbers-less-than-20-is-closest-126587.html

Hope it helps.



Is there a more logical way of solving?
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Re: The product of the first eight positive even integers is closest to  [#permalink]

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New post 22 Nov 2013, 07:49
1
I solved it this way... since this is approximation question, we can approximate as below

\(2*4*6*8*10*12*14*16\) rounding up or down each of the numbers to the nearest 10.

\(2*4*10*10*10*10*10*20\)
= \(2*4*10^5*2*10\)
= \(16*10^6\)
=\(20*10^6\)
=\(2*10^7\)

it is closer to \(10^7\) than \(10^8\). so its C.
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Re: The product of the first eight positive even integers is closest to  [#permalink]

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New post 28 Jan 2017, 02:37
I solved it the following way:
2*4*6*8*10*12*14*16
= 1*5*5*10*10*10*15*20 (approximately)
=25*1000*300
=75 *10^5
=100*10^5 (approximately)
=10^7
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Re: The product of the first eight positive even integers is closest to  [#permalink]

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New post 18 Aug 2018, 05:20
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Re: The product of the first eight positive even integers is closest to &nbs [#permalink] 18 Aug 2018, 05:20
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