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If the 4 x 4 grid pictured at right is filled with the consecutive

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enigma123
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If the 4 x 4 grid pictured at right is filled with the consecutive [#permalink] New post   31 Jan 2012, 19:25
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If the 4 x 4 grid pictured at right is filled with the consecutive integers from 37 to 52, inclusive, so that every row, column and major diagonal sums to the same value, which of the following is a possible value of the sum of the four center cells of the grid (indicated by the shaded area)?

(A) 124
(B) 153
(C) 178
(D) 192
(E) 214

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C
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Re: If the 4 x 4 grid pictured at right is filled with the consecutive [#permalink] New post   01 Feb 2012, 03:58
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Sum of the numbers from 37 to 52 = (n/2)[2a + (n-1)d] = (16/2) [74+15] = 8 * 89 = 712

Therefore the sum of all numbers in the grid = 712
If x is the sum of all numbers in a row/column/major diagonal, then
4x = 712
=> x = 178 = sum of all numbers in any row = sum of all numbers in any diagonal = sum of all numbers in either major diagonal

Now, consider the grid as follows:
1' 2' 3' 4'
5' 6' 7' 8'
9' 10' 11' 12'
13' 14' 15' 16'

We know that 1' + 6' + 11' + 16' = 178
4' + 7' + 10' + 13' = 178
=> 1' + 6' + 11' + 16' + 4' + 7' + 10' + 13' = 356
=> 6' + 7' + 10' + 11' + 1' + 13' + 4' + 16' = 356

Also 5' + 6' + 7' + 8' + 9' + 10' + 11' + 12' = 356
and 1' + 5' + 9' + 13' + 4' + 8' + 12' + 16' = 356
=> 6' + 7' + 10' + 11' - 1' - 13' - 4' - 16' = 0

Therefore 6' + 7' + 10' + 11' = 356/2 = 178 = sum of the middle four numbers

Option (C).
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Re: If the 4 x 4 grid pictured at right is filled with the consecutive [#permalink] New post   13 Jul 2016, 12:04
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enigma123 wrote:
If the 4 x 4 grid in the attached picture is filled with the consecutive integers from 37 to 52, inclusive, so that every row, column and major diagonal sums to the same value, which of the following is a possible value of the sum of the four center cells of the grid (indicated by the shaded area)?

(A) 124
(B) 153
(C) 178
(D) 192
(E) 214

Any idea how to approach this question please?


I approached it this way:

there are 16 cells; 37,38,39,...,52 can be denoted as 36+1,36+2,36+3,...,36+16
the total of 37+38+...+52 is therefore

=36*16+16*17/2
=16(36+8.5)
=16*44.5
=8*89
=4*178
=2*356
=712

The average value of a cell is 712/16 = 44.5

Attachment:
gmatclub.PNG
gmatclub.PNG [ 1.95 KiB | Viewed 14553 times ]


Also given : sum of values in column 2 = sum of values in column 3 = sum of values in row 2 = sum of values in row 3.

To satisfy this, if we assigned the average value of a cell to each cell, the 4 cells will get a total of 44.5*4 = 178.

Hence C is the answer.
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Re: If the 4 x 4 grid pictured at right is filled with the consecutive [#permalink] New post   08 Dec 2013, 08:01
In the last step "=> 6' + 7' + 10' + 11' - 1' - 13' - 4' - 16' = 0"
how can we be sure that the +ve groupings and -ve groupings each add up to 178 or 6' + 7' + 10' + 11' = 178. It could be 180 for 6' + 7' + 10' + 11' and 176 for - 1' - 13' - 4' - 16'. Am i missing something here?
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Re: If the 4 x 4 grid pictured at right is filled with the consecutive [#permalink] New post   08 Dec 2013, 08:02
In the last step "=> 6' + 7' + 10' + 11' - 1' - 13' - 4' - 16' = 0"
how can we be sure that the +ve groupings and -ve groupings each add up to 178 or 6' + 7' + 10' + 11' = 178. It could be 180 for 6' + 7' + 10' + 11' and 176 for - 1' - 13' - 4' - 16'. Am i missing something here?
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Re: If the 4 x 4 grid pictured at right is filled with the consecutive [#permalink] New post   04 Jan 2016, 19:57
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I followed a rather simple approach. Since all the four sides and the major diagonal has same value. The sum of four center cell should also be the same. and it should be (37+38+51+52) = 178
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Re: If the 4 x 4 grid pictured at right is filled with the consecutive [#permalink] New post   13 Jul 2016, 22:31
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mean of all 16 numbers is 44.5
take two number on right of this mean and take two number left , but they all are in such a way that mean is 44.5.
so basically 44.5* 4 = 178
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Re: If the 4 x 4 grid pictured at right is filled with the consecutive [#permalink] New post   29 Dec 2016, 20:16
I solved it in this way:

Min number is 37 so Min Sum of 4 numbers will be 37*4 = 148. (Option A out).
Max Number is 52 so max sum of 4 numbers will be 52*4 = 208 (Option E out).

Because the numbers are uniformly distributed their sum should should lie in middle of 144 and 208 not towards any of these numbers.(Most likely B and D are out.)
148+208 = 356/2 = 178.

Hope this logic makes sense.
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Re: If the 4 x 4 grid pictured at right is filled with the consecutive [#permalink] New post   19 May 2017, 22:05
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hello .

i blv we can solve it using algebra.
sum of n numbers as we know
Sn =n/2(2a +(n-1)d Here d=1, n=16, a=37

we will get that sum of all from 37-52 incl is 712.
but we have to find sum in each column & dignoal uniform so =712/4 =178 hence ans is C
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Re: If the 4 x 4 grid pictured at right is filled with the consecutive [#permalink] New post   19 Jan 2018, 03:29
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enigma123 wrote:
If the 4 x 4 grid in the attached picture is filled with the consecutive integers from 37 to 52, inclusive, so that every row, column and major diagonal sums to the same value, which of the following is a possible value of the sum of the four center cells of the grid (indicated by the shaded area)?

(A) 124
(B) 153
(C) 178
(D) 192
(E) 214

Any idea how to approach this question please?


Here's how I solved it

the number of integers= 52-37+1=16
The median (8th & 7th term)= (37+8)+(37+7)= 44&45 (No need to divide by 2)
Add the previous number and the following number= 43+44+45+46= 178 (D)
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Re: If the 4 x 4 grid pictured at right is filled with the consecutive [#permalink] New post   19 Jan 2018, 04:11
manojach87 wrote:
enigma123 wrote:
If the 4 x 4 grid in the attached picture is filled with the consecutive integers from 37 to 52, inclusive, so that every row, column and major diagonal sums to the same value, which of the following is a possible value of the sum of the four center cells of the grid (indicated by the shaded area)?

(A) 124
(B) 153
(C) 178
(D) 192
(E) 214

Any idea how to approach this question please?


I approached it this way:

there are 16 cells; 37,38,39,...,52 can be denoted as 36+1,36+2,36+3,...,36+16
the total of 37+38+...+52 is therefore

=36*16+16*17/2
=16(36+8.5)
=16*44.5
=8*89
=4*178
=2*356
=712

The average value of a cell is 712/16 = 44.5

Attachment:
gmatclub.PNG


Also given : sum of values in column 2 = sum of values in column 3 = sum of values in row 2 = sum of values in row 3.

To satisfy this, if we assigned the average value of a cell to each cell, the 4 cells will get a total of 44.5*4 = 178.

Hence C is the answer.


I believe there's a simpler way to calculate the total:

[#of items (first term+last term)] /2 = 16 (37+52)/2= 8*89= 712
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Re: If the 4 x 4 grid pictured at right is filled with the consecutive [#permalink] New post   30 Jun 2018, 11:26
enigma123 wrote:
If the 4 x 4 grid in the attached picture is filled with the consecutive integers from 37 to 52, inclusive, so that every row, column and major diagonal sums to the same value, which of the following is a possible value of the sum of the four center cells of the grid (indicated by the shaded area)?

(A) 124
(B) 153
(C) 178
(D) 192
(E) 214

Any idea how to approach this question please?

10-If the 4 x 4 grid in the attached picture is filled with the consecutive integers from 37 to 52, inclusive, so that every row, column and major diagonal sums to the same value, which of the following is a possible value of the sum of the four center cells of the grid (indicated by the shaded area)?

(A) 124
(B) 153
(C) 178
(D) 192
(E) 214


37 38 39 40 178
44 43 42 41 178
45 46 47 48 178
52 51 50 49 178
178 178 178 178

C
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Re: If the 4 x 4 grid pictured at right is filled with the consecutive [#permalink] New post   30 Jun 2018, 21:07
enigma123 wrote:
If the 4 x 4 grid in the attached picture is filled with the consecutive integers from 37 to 52, inclusive, so that every row, column and major diagonal sums to the same value, which of the following is a possible value of the sum of the four center cells of the grid (indicated by the shaded area)?

(A) 124
(B) 153
(C) 178
(D) 192
(E) 214

Any idea how to approach this question please?


The Sum will be two of the least number (37+38) + tow of the max (51 + 52) = 178
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Re: If the 4 x 4 grid pictured at right is filled with the consecutive [#permalink] New post   22 Jan 2023, 04:10
GyanOne wrote:
Sum of the numbers from 37 to 52 = (n/2)[2a + (n-1)d] = (16/2) [74+15] = 8 * 89 = 712

Therefore the sum of all numbers in the grid = 712
If x is the sum of all numbers in a row/column/major


diagonal, then
4x = 712
=> x = 178 = sum of all numbers in any row = sum of all numbers in any diagonal = sum of all numbers in either major diagonal

Now, consider the grid as follows:
1' 2' 3' 4'
5' 6' 7' 8'
9' 10' 11' 12'
13' 14' 15' 16'

We know that 1' + 6' + 11' + 16' = 178
4' + 7' + 10' + 13' = 178
=> 1' + 6' + 11' + 16' + 4' + 7' + 10' + 13' = 356
=> 6' + 7' + 10' + 11' + 1' + 13' + 4' + 16' = 356

Also 5' + 6' + 7' + 8' + 9' + 10' + 11' + 12' = 356
and 1' + 5' + 9' + 13' + 4' + 8' + 12' + 16' = 356
=> 6' + 7' + 10' + 11' - 1' - 13' - 4' - 16' = 0

Therefore 6' + 7' + 10' + 11' = 356/2 = 178 = sum of the middle four numbers

Option (C).



Hi please advice if the as per question stem each consecutive box will be filled with consecutive integer starting from 37-52

37,36,38,39,40
41,42,43,44,45 and so on

Or we will use consecutive integers from 37-52 but they can be in any box in 4*4 grid

Posted from my mobile device
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Re: If the 4 x 4 grid pictured at right is filled with the consecutive [#permalink] New post   08 Sep 2023, 10:24
enigma123 wrote:
If the 4 x 4 grid in the attached picture is filled with the consecutive integers from 37 to 52, inclusive, so that every row, column and major diagonal sums to the same value, which of the following is a possible value of the sum of the four center cells of the grid (indicated by the shaded area)?

(A) 124
(B) 153
(C) 178
(D) 192
(E) 214



If you want every row, column and diagonal to sum to same value, the middle needs to be in the centre.
Middle numbers of the set : 43,44,45,46

Unit digits of 3+4+5+6 sums of 8 in units place.
Hence C
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Re: If the 4 x 4 grid pictured at right is filled with the consecutive [#permalink] New post   09 Oct 2023, 07:03
KarishmaB - Is this a GMAT like question? Can you kindly explain what really is the concept involved as random testing of numbers to make them equal took a lot of time?

Thank you
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Re: If the 4 x 4 grid pictured at right is filled with the consecutive [#permalink] New post   10 Oct 2023, 02:43
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Sonia0106 wrote:
KarishmaB - Is this a GMAT like question? Can you kindly explain what really is the concept involved as random testing of numbers to make them equal took a lot of time?

Thank you


GMAT could give some puzzle type of questions though they would have logic in them. It wouldn't give you a question in which you need to use hit and trial and solve for exact values. You may not be able to neatly bracket them into one topic.
Here I would just find the average and take it 4 times. The number distribution is symmetrical - all rows, columns and diagonals have the same sum so likely the centre four numbers would have the same sum too.
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