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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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Updated on: 26 Dec 2017, 09:37
4
00:00

Difficulty:

75% (hard)

Question Stats:

55% (02:04) correct 45% (01:59) wrong based on 320 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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25 Feb 2012, 17:16
3
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).

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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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).

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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Joined: 25 Oct 2013
Posts: 143
Re: The product of the first eight positive even integers is closest to  [#permalink]

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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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28 Jan 2017, 02:37
1
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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09 Jan 2019, 07:37
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
Re: The product of the first eight positive even integers is closest to   [#permalink] 09 Jan 2019, 07:37
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