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

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Math Expert
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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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29 Apr 2015, 04:02
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Difficulty:

75% (hard)

Question Stats:

61% (02:51) correct 39% (02:46) wrong based on 93 sessions

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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

Attachment:

Screen-Shot-2012-05-18-at-10.58.53-PM.png [ 8.69 KiB | Viewed 1570 times ]

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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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Updated on: 07 May 2015, 00:12
2
3
(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.

Originally posted by PrepTap on 29 Apr 2015, 05:49.
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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29 Apr 2015, 05:32
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Bunuel wrote:

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

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

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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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01 May 2015, 00:22
1
(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.

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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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04 May 2015, 04:26
Bunuel wrote:

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

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:

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.

Attachment:

Screen-Shot-2012-05-18-at-11.03.10-PM.png [ 7.26 KiB | Viewed 1277 times ]

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