experts help plz

[quantskillsgmat]

in a grid of 4x4, consecutive integers from 38 to 52 inclusive have to be arranged in such a way so that every column, row and major diagonals sum to same value. which of the following is a possible value of sum of four central cells.
a)124 b)153 c)178 d)192 d)214

[sanju09]

in a grid of 4x4, consecutive integers from 38 to 52 inclusive have to be arranged in such a way so that every column, row and major diagonals sum to same value. which of the following is a possible value of sum of four central cells.
a)124 b)153 c)178 d)192 d)214

Make a small correction in the question; take consecutive integers from 37 to 52 inclusive, instead of 38 to 52.

This is a question frequently played in leisure mathematics, where there is a magic square of order n in which n^2 distinct integers are to be arranged, such that the n integers in all rows, all columns, and in both major diagonals sum to the same constant, which is called the magic constant or magic sum.

A typical magic square contains the integers from 1 to n^2 and the magic sum depends on the value of n only. The magic sum is equal to ½ n (n^2 + 1), and is same as sum of n central cells as well.

Hence, a 4 × 4 magic square will have a magic sum of ½ (4) (17) = 34. If the integers are 37 to 52, it means all the 4 integers in the 4 central cells are also increased by 36. Hence there will be a net increase of 4 × 36 = 144 in the magic sum 34, to get 178 as the answer to the question presented.

Take C

[GMATGuruNY]

in a grid of 4x4, consecutive integers from 38 to 52 inclusive have to be arranged in such a way so that every column, row and major diagonals sum to same value. which of the following is a possible value of sum of four central cells.
a)124 b)153 c)178 d)192 d)214

Let the grid look as follows:
ABCD
EFGH
IJKL
MNOP

The grid must contain 16 integers.
Let's assume that the intended range is from 37 to 52.

Given consecutive integers:
The number of integers = biggest-smallest+1 = 52-37+1 = 16.
The average of the integers = (biggest+smallest)/2 = (57+32)/2 = 89/2.
The sum of the integers = number*average = 16*(89/2) = 712.

Thus, the sum of the entire grid must be 712.
Since the sum of the integers in each row must be the same, each row = 712/4 = 178.
Thus, the sum of the integers in any row, column, or main diagonal = 178.

We can plug in the answers, which represent the sum of the four central cells.
It seems VERY likely that the correct answer will be C:
F+G+J+K=178.
Since the sum of rows 2 and 3 must be 2*178, E+H+I+L=178.
Since the sum of columns 2 and 3 must be 2*178, B+C+N+O =178.
Since the sum of the two main diagonals must be 2*178, A+D+M+P=178.
These 4 sums represent the entire grid.
Thus, the sum of the entire grid = 4*178=712.
Success!
If we plug in any other answer choice, the sum of the entire grid will not be 712.

The correct answer is C.

[quantskillsgmat]

thanx mitch

[sanju09]

(7 + 36) (12 + 36) (1 + 36) (14 + 36)
(2 + 36) (13 + 36) (8 + 36) (11 + 36)
(16 + 36) (3 + 36) (10 + 36) (5 + 36)
(9 + 36) (6 + 36) (15 + 36) (4 + 36)


The question asks about the sum of four central cells. Such a question about magic square of order n is possible to make only if n is itself a perfect square like 4, 9, 16, etc. We can't presume the 2nd or the 3rd row of a 4 × 4 magic square as four central cells. Hence, this fact about the magic square of order n, where n is a perfect square integer equal to or more than 4, is indispensable to know that the sum of n central cells is same as the magic sum ½ n (n^2 + 1).

[sanju09]

in a grid of 4x4, consecutive integers from 38 to 52 inclusive have to be arranged in such a way so that every column, row and major diagonals sum to same value. which of the following is a possible value of sum of four central cells.
a)124 b)153 c)178 d)192 d)214

Let the grid look as follows:
ABCD
EFGH
IJKL
MNOP

The grid must contain 16 integers.
Let's assume that the intended range is from 37 to 52.

Given consecutive integers:
The number of integers = biggest-smallest+1 = 52-37+1 = 16.
The average of the integers = (biggest+smallest)/2 = (57+32)/2 = 89/2.
The sum of the integers = number*average = 16*(89/2) = 712.

Thus, the sum of the entire grid must be 712.
Since the sum of the integers in each row must be the same, each row = 712/4 = 178.
Thus, the sum of the integers in any row, column, or main diagonal = 178.

We can plug in the answers, which represent the sum of the four central cells.
It seems VERY likely that the correct answer will be C:
F+G+J+K=178.
Since the sum of rows 2 and 3 must be 2*178, E+H+I+L=178.
Since the sum of columns 2 and 3 must be 2*178, B+C+N+O =178.
Since the sum of the two main diagonals must be 2*178, A+D+M+P=178.
These 4 sums represent the entire grid.
Thus, the sum of the entire grid = 4*178=712.
Success!
If we plug in any other answer choice, the sum of the entire grid will not be 712.

The correct answer is C.

Hi Mitch, I didn't get the logic behind all what's in bold in your explanation above. How's plug-in working here? Please enlighten.

Best regards

[GMATGuruNY]

in a grid of 4x4, consecutive integers from 38 to 52 inclusive have to be arranged in such a way so that every column, row and major diagonals sum to same value. which of the following is a possible value of sum of four central cells.
a)124 b)153 c)178 d)192 d)214

Let the grid look as follows:
ABCD
EFGH
IJKL
MNOP

The grid must contain 16 integers.
Let's assume that the intended range is from 37 to 52.

Given consecutive integers:
The number of integers = biggest-smallest+1 = 52-37+1 = 16.
The average of the integers = (biggest+smallest)/2 = (57+32)/2 = 89/2.
The sum of the integers = number*average = 16*(89/2) = 712.

Thus, the sum of the entire grid must be 712.
Since the sum of the integers in each row must be the same, each row = 712/4 = 178.
Thus, the sum of the integers in any row, column, or main diagonal = 178.

We can plug in the answers, which represent the sum of the four central cells.
It seems VERY likely that the correct answer will be C:
F+G+J+K=178.
Since the sum of rows 2 and 3 must be 2*178, E+H+I+L=178.
Since the sum of columns 2 and 3 must be 2*178, B+C+N+O =178.
Since the sum of the two main diagonals must be 2*178, A+D+M+P=178.
These 4 sums represent the entire grid.
Thus, the sum of the entire grid = 4*178=712.
Success!
If we plug in any other answer choice, the sum of the entire grid will not be 712.

The correct answer is C.

Hi Mitch, I didn't get the logic behind all what's in bold in your explanation above. How's plug-in working here? Please enlighten.

Best regards

The grid looks as follows:
ABCD
EFGH
IJKL
MNOP
The sum of the integers in any row, column, or main diagonal = 178.

According to answer choice C, F+G+J+K=178.
The sum of the integers in row 2 must be 178: E+F+G+H=178
The sum of the integers in row 3 must be 178: I+J+K+L=178.
Thus, the sum of row 2 and row 3 = (E+F+G+H) + (I+J+K+L) = 2*178.
Subtracting F+G+J+K=178, we get:
E+H+I+L=178.

I applied the same reasoning to calculate that B+C+N+O=178 and that A+D+M+P=178.

[tomada]

Did this question come from a GMATPrep test?

[tomada]

I like Mitch's solution. Reminds me of linear algebra