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In the grid above, the variables a through p are each equal to 2, 3, 5

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In the grid above, the variables a through p are each equal to 2, 3, 5 [#permalink]

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In the grid above, the variables a through p are each equal to 2, 3, 5, or 7, with exactly one occurrence of each value in any row and in any column. What is the value of fgjk?

(1) bcehilno = 35^4

(2) (2.25^2)afkp = dgjm


[Reveal] Spoiler:
Attachment:
Screen-Shot-2012-05-18-at-10.58.53-PM.png
Screen-Shot-2012-05-18-at-10.58.53-PM.png [ 8.69 KiB | Viewed 1332 times ]
[Reveal] Spoiler: OA

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In the grid above, the variables a through p are each equal to 2, 3, 5 [#permalink]

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New post 29 Apr 2015, 05:32
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Bunuel wrote:
Image
In the grid above, the variables a through p are each equal to 2, 3, 5, or 7, with exactly one occurrence of each value in any row and in any column. What is the value of fgjk?

(1) bcehilno = 35^4

(2) (2.25^2)afkp = dgjm


[Reveal] Spoiler:
Attachment:
Screen-Shot-2012-05-18-at-10.58.53-PM.png


1) \(35^4 = 7^4*5^4\) So we can infer that all letters bcehilno are equal to \(5\) or \(7\)
and we see on the picture that \(fj\) placed in vertical row with \(b\) and \(n\) so it should be equal to \(2\) and \(3\)
and \(gk\) placed in vertical row with \(c\) and \(o\) so it should be equal to \(2\) and \(3\)
We can infer that \(fgjk\) will be equal to \(2*3*2*3=36\)
Sufficient

2) This is diagonal letters, so it should be equal numbers like \(2222\) or \(3333\) etc.
we have 4 variants: \(2^4 = 16\); \(3^4=81\); \(5^4=625\); \(7^4=\)something big, we don't need it
\(2.25^2\) equal to \(5\) and something more
and we see that \(2.25^2*2^4=3^4=81\) (I didn't check it but 5*16 = 80 and we have something more in 5 so it one possible variant I think)
so we can infer that \(fk = 2*2\) and \(gj=3*3\)
\(fgjk\) will be equal to \(2*3*2*3=36\)
Sufficient

Answer is D
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In the grid above, the variables a through p are each equal to 2, 3, 5 [#permalink]

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(1) \(bcehilno = 35^4 = 5^4 * 7^4\)

This means 4 out of the 8 numbers b,c,e,h,i,l,n,o are equal to 5 and the rest 4 are equal to 7.

Since e and h are in the same row, one of them is 5 and the other one 7.
Therefore, f and g are either 2 and 3 or 3 and 2 respectively.
Similarly, j and k are 2 and 3 or 3 and 2 respectively.

So f, g, j, k are 2, 3, 3, 2 or 3, 2, 2, 3 respectively since numbers in the same row or column can't have the same value.
In both cases, \(fjgk = 2*3*3*2 = 36\)
SUFFICIENT

(2) (2.25^2)afkp = dgjm

2.25 can also be written as \(\frac{9}{4}\).

\(9^2 afkp = 4^2 dgjm\)
\(3^4 afkp = 2^4 dgjm\)

Since 3 and 2 are prime numbers, the only way LHS and RHS can be equal if \(afkp = 2^4\) and \(dgjm = 3^4\)
Therefore, a = f = k = p = 2 and d = g = j = m = 3
\(fjgk = 2*3*3*2 = 36\)
SUFFICIENT

D is the correct choice here.

Last edited by PrepTap on 07 May 2015, 00:12, edited 1 time in total.

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In the grid above, the variables a through p are each equal to 2, 3, 5 [#permalink]

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(1) bcehilno = \(35^4 = 5^4 * 7^4\)
so bc=35; eh=35 ,il=35 , no=35
hence fg=2*3=6 and kj=2*3=6

Sufficient.


(2) (2.25^2)afkp = dgjm
\(2.25^2 = 1.5*1.5*1.5*1.5 = \frac{3}{2}* \frac{3}{2}* \frac{3}{2}* \frac{3}{2}\)
so, d,g,j,m=3 and a,f,k,p=2
sufficient.

Answer D
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Re: In the grid above, the variables a through p are each equal to 2, 3, 5 [#permalink]

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Bunuel wrote:
Image
In the grid above, the variables a through p are each equal to 2, 3, 5, or 7, with exactly one occurrence of each value in any row and in any column. What is the value of fgjk?

(1) bcehilno = 35^4

(2) (2.25^2)afkp = dgjm


[Reveal] Spoiler:
Attachment:
The attachment Screen-Shot-2012-05-18-at-10.58.53-PM.png is no longer available


MANHATTAN GMAT OFFICIAL SOLUTION:

Although there are 16 different variables in this 4×4 grid, the values of these variables are severely restricted: they can only be 2, 3, 5, or 7 (you should recognize these as primes), and only one of each can exist in any row or column. You are asked for the value of the product fgjk, which is the 2×2 grid at the center.

(1) SUFFICIENT. Note that 35^4 = 5^4*7^4, so four of the variables must be 5’s, while the other four must be 7’s. You are not allowed to put two of the same values in any row or column, so a possible arrangement (there are many) looks like this:
Image

The central grid must be composed of two 2’s and two 3’s, since each row and each column must have exactly one of each possible value, and you’ve already used up the 5’s and 7’s. Although you cannot be sure which particular variable in the central 2×2 grid is which value, you know that the product is 36 (2×2×3×3).

(2) SUFFICIENT. First, use FDP connections: rewrite 2.25 as 9/4, or (3/2)^2. Thus, 2.25^2= (3/2)^4. Cross-multiplying by 2^4, you get \(3^4afkp = 2^4dgjm\).

Since the only possible values of the variables are 2, 3, 5, and 7, the only possibilities for the values of the variables are a = f = k = p = 2, while d = g = j = m = 3.

This tells us that fgjk = 36.

The correct answer is D.

[Reveal] Spoiler:
Attachment:
Screen-Shot-2012-05-18-at-11.03.10-PM.png
Screen-Shot-2012-05-18-at-11.03.10-PM.png [ 7.26 KiB | Viewed 1052 times ]

_________________

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Resources:
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Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


What are GMAT Club Tests?
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Re: In the grid above, the variables a through p are each equal to 2, 3, 5 [#permalink]

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Re: In the grid above, the variables a through p are each equal to 2, 3, 5   [#permalink] 17 Oct 2017, 20:51
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