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The table above is a 3 by 3 grid of 9 numbers, 5 of which are given
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Updated on: 30 Jul 2016, 06:16
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The table above is a 3 by 3 grid of 9 numbers, 5 of which are given. The 4 numbers that are not given are denoted by A, B, C, and D. The product of the numbers in each row and in each column is the same for all rows and columns. What is the value of the product ABCD ? A) 36 B) 72 C) \(36\sqrt{2}\) D) \(36\sqrt{3}\) E) \(72\sqrt{6}\) Attachment:
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Originally posted by NandishSS on 30 Jul 2016, 05:05.
Last edited by Bunuel on 30 Jul 2016, 06:16, edited 1 time in total.
Edited the question.




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Re: The table above is a 3 by 3 grid of 9 numbers, 5 of which are given
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30 Jul 2016, 07:42
A * B * sqrt(3) = 6*sqrt(6) A * B = 6 * sqrt(2)
C * D * 2sqrt(3) = 6*sqrt(6) C * D = 3 * sqrt(2)
A * B * C * D = 6 * 3 * 2 = 36
Answer: A




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Re: The table above is a 3 by 3 grid of 9 numbers, 5 of which are given
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14 Aug 2016, 09:50
Hi all,
Does anyone know where 6*sqrt(6) is coming from?
Thank you.



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Re: The table above is a 3 by 3 grid of 9 numbers, 5 of which are given
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14 Aug 2016, 19:48
HarveyKlaus wrote: Hi all,
Does anyone know where 6*sqrt(6) is coming from?
Thank you. Refer to the center column that has numbers 1, 6 and sqrt(6). Multiply and you will get 6*sqrt(6).



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Re: The table above is a 3 by 3 grid of 9 numbers, 5 of which are given
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22 Jun 2017, 07:40
just square each number in the table and then the solution becomes obvious A 1 12 B 6 6 3 6 D so we need 6*6=36 cheers
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Re: The table above is a 3 by 3 grid of 9 numbers, 5 of which are given
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22 Jun 2017, 11:32
daviddaviddavid wrote: just square each number in the table and then the solution becomes obvious
A 1 12 B 6 6 3 6 D
so we need 6*6=36
cheers 6^2 does not equal 6, (see column 2, row 3). Can you elaborate?
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The table above is a 3 by 3 grid of 9 numbers, 5 of which are given
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22 Jun 2017, 12:24
Smokeybear00 wrote: daviddaviddavid wrote: just square each number in the table and then the solution becomes obvious
A 1 12 B 6 6 3 6 D
so we need 6*6=36
cheers 6^2 does not equal 6, (see column 2, row 3). Can you elaborate? If you look at the question, the middle column reads 1,6,\(\sqrt{6}\) not 1,6,6 Hence squaring it will give you 1,36 and 6. Btw, why are you squaring the numbers? The solution to this problem is as follows :
Since the product of each of the rows and columns must be equal, only the center row has all three values, the product of which is 6*\(\sqrt{6}\) We need to find the product of A,B,C,D. From the first row : A * B * sqrt(3) = 6*sqrt(6) ==> A * B = 6 * sqrt(2) From the third row :C * D * 2sqrt(3) = 6*sqrt(6) ==> C * D = 3 * sqrt(2) Therefore product of A,B,C and D as asked in the question is (6*sqrt(2))*(3*sqrt(2)) =6*3*2 = 36 (Option A)Hope it helps!
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Re: The table above is a 3 by 3 grid of 9 numbers, 5 of which are given
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23 Jun 2017, 00:40
Smokeybear00 wrote: daviddaviddavid wrote: just square each number in the table and then the solution becomes obvious
A 1 12 B 6 6 3 6 D
so we need 6*6=36
cheers 6^2 does not equal 6, (see column 2, row 3). Can you elaborate? well, Im sorry I made a careless mistake, but you should get the point A 1 12 B 6 6 3 36 D
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Re: The table above is a 3 by 3 grid of 9 numbers, 5 of which are given
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23 Jun 2017, 00:41
pushpitkc wrote: Smokeybear00 wrote: daviddaviddavid wrote: just square each number in the table and then the solution becomes obvious
A 1 12 B 6 6 3 6 D
so we need 6*6=36
cheers 6^2 does not equal 6, (see column 2, row 3). Can you elaborate? If you look at the question, the middle column reads 1,6,\(\sqrt{6}\) not 1,6,6 Hence squaring it will give you 1,36 and 6. Btw, why are you squaring the numbers? The solution to this problem is as follows :
Since the product of each of the rows and columns must be equal, only the center row has all three values, the product of which is 6*\(\sqrt{6}\) We need to find the product of A,B,C,D. From the first row : A * B * sqrt(3) = 6*sqrt(6) ==> A * B = 6 * sqrt(2) From the third row :C * D * 2sqrt(3) = 6*sqrt(6) ==> C * D = 3 * sqrt(2) Therefore product of A,B,C and D as asked in the question is (6*sqrt(2))*(3*sqrt(2)) =6*3*2 = 36 (Option A)Hope it helps! yes thanks, I made a careless mistake I squared the numbers, because for me the problem becomes much easier
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Re: The table above is a 3 by 3 grid of 9 numbers, 5 of which are given
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26 Jun 2017, 17:01
NandishSS wrote: The table above is a 3 by 3 grid of 9 numbers, 5 of which are given. The 4 numbers that are not given are denoted by A, B, C, and D. The product of the numbers in each row and in each column is the same for all rows and columns. What is the value of the product ABCD ? A) 36 B) 72 C) \(36\sqrt{2}\) D) \(36\sqrt{3}\) E) \(72\sqrt{6}\) We can start with the middle column: 1 x √6 x 6 = 6√6. Thus, we know that the entries in each row and in each column multiply to yield 6√6. For Row 1: A x 1 x 2√3 = 6√6 A = (6√6)/(2√3) A = 3√2 For Column 1: 3√2 x B x √3 = 6√6 B x 3√6 = 6√6 B = (6√6)/(3√6) B = 2 For Row 2: 2 x √6 x C = 6√6 C = (6√6)/(2√6) C = 3 For Row 3: √3 x 6 x D = 6√6 D = (6√6)/(6√3) D = √2 Thus, ABCD = 3√2 x 2 x 3 x √2 = 18√2 x √2 = 36 Answer: A
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Re: The table above is a 3 by 3 grid of 9 numbers, 5 of which are given
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27 Jun 2017, 13:49
The table above is a 3 by 3 grid of 9 numbers, 5 of which are given. The 4 numbers that are not given are denoted by A, B, C, and D. The product of the numbers in each row and in each column is the same for all rows and columns. What is the value of the product ABCD ? From Second column we see that the multiplication for each row/columns should be \(= 6 * \sqrt{6}\) \(= 2 * 3 * \sqrt{3} * \sqrt{2}\) \(AB * \sqrt{3}\) \(= 2 * 3 * \sqrt{3} * \sqrt{2}\) \(AB = 6 * \sqrt{2}\) \(CD * 2\sqrt{3}\) \(= 2 * 3 * \sqrt{3} * \sqrt{2}\) \(CD = 3 * \sqrt{2}\) \(AB * CD = 6 * \sqrt{2} * 3 * \sqrt{2}\) \(AB * CD = 18 * 2\) \(AB * CD = 36\) Hence, Answer is A
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Re: The table above is a 3 by 3 grid of 9 numbers, 5 of which are given
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15 Jul 2018, 06:56
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