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The product of the first seven positive multiples of three is closest

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The product of the first seven positive multiples of three is closest  [#permalink]

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New post 10 Jan 2012, 08:28
3
00:00
A
B
C
D
E

Difficulty:

  65% (hard)

Question Stats:

53% (01:52) correct 47% (01:55) wrong based on 196 sessions

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The product of the first seven positive multiples of three is closest to which of the following powers of 10?

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

How to solve it in less than two minutes?

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The product of the first seven positive multiples of three is closest  [#permalink]

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New post 04 Jan 2018, 15:19
3
solved using approximation may not always work

(3, 6) = 18
(9) = 10 (approx)
(12, 15) = 180
(18, 21) = (20 x 20) = 400 approx

180 * 180 * 400 = 324 * 4 * 10000
still approximating = 300 * 4 * 10000 = 12000000 = approx
closest is \(10^7\)
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Re: The product of the first seven positive multiples of three is closest  [#permalink]

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New post 10 Jan 2012, 21:21
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subhajeet wrote:
I could not understand the question. Can you please help me on that. According to me the product must be only 3^7.


Multiples of 3: 3*-2, 3*-1, 3*1, 3*2, 3*3 and so on....
first seven positive multiples of three: 3, 6, 9, 12, 15, 18, and 21

Multiply them all, and you will get (since each of the seven multiples have one '3'): 3^7 * (1*2*3*4*5*6*7) = (3^7)*(7!)
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Re: The product of the first seven positive multiples of three is closest  [#permalink]

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New post 10 Jan 2012, 11:44
The product will be 3^7 * 1 * 2 * 3 * 4 * 5 * 6 * 7
=3^7 * 5140
= 9^3 * 15420
= ~135000 * 9 * 9
= ~1220000 * 9
= ~11000000 which is very close to 10^7

The correct answer is therefore (C)
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Re: The product of the first seven positive multiples of three is closest  [#permalink]

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New post 10 Jan 2012, 21:07
GyanOne wrote:
The product will be 3^7 * 1 * 2 * 3 * 4 * 5 * 6 * 7
=3^7 * 5140
= 9^3 * 15420
= ~135000 * 9 * 9
= ~1220000 * 9
= ~11000000 which is very close to 10^7

The correct answer is therefore (C)


I could not understand the question. Can you please help me on that. According to me the product must be only 3^7.
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Re: The product of the first seven positive multiples of three is closest  [#permalink]

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New post 10 Jan 2012, 22:56
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1
I took a bit of a gamble.

I quickly took the first seven positive multiples of 3:

3
6
9
12
15
18
21

At a quick glance, I determined that the final product would probably be a seven-digit number. Hence, C.
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Re: The product of the first seven positive multiples of three is closest  [#permalink]

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New post 09 Jan 2019, 07:48
Write out the multiples: 3, 6, 9, 12, 15, 18, 21
Pick from the numbers to get as many products close to powers of 10 as possible

9*12 = 108 =~ 100
6*15 = 90 =~ 100
18*21, almost the same as 20*20 so 4*100
left with 3, so 12*10^6 --> 1.2*10^7

P.S. Actual number is 11022480, so very close to 1.1*10^7
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Re: The product of the first seven positive multiples of three is closest  [#permalink]

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New post 12 Jan 2019, 07:50
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3*6*9*12*15*18*21= 3^7*7!

3^7= 729*3= approx 2100
7!= 720*7= approx 5000
Approx Product= 10^7
Ans C
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Re: The product of the first seven positive multiples of three is closest   [#permalink] 12 Jan 2019, 07:50
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