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655-705 (Hard)|   Arithmetic|      
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enigma123
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The product of the first eight positive even integers is closest to Updated on: 26 Dec 2017, 09:37
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.
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C
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65% (hard)

Question Stats:

55% (02:04) correct 45% (02:00) wrong based on 541 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) \(10^5\)

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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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C
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Bunuel
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The product of the first eight positive even integers is closest to 25 Feb 2012, 17:16
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enigma123
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?
Show more

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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The product of the first eight positive even integers is closest to 22 Nov 2013, 07:29
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Bunuel
enigma123
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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Is there a more logical way of solving?
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gmatprav
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The product of the first eight positive even integers is closest to 22 Nov 2013, 07:49
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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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The product of the first eight positive even integers is closest to 28 Jan 2017, 02:37
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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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The product of the first eight positive even integers is closest to 09 Jan 2019, 07:37
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The goal is to pick from the positive even ints 2,4,6,8,10,12,14,16 to get products close to powers of 10.

10
6*16=96 =~ 100
8*12=96 =~ 100
4*14*2 = 56*2 = 112 =~ 100

So, we get something close to 10^7
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The product of the first eight positive even integers is closest to 29 Jan 2025, 04:04
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