16 numbers (not necessarily distinct) are filled in the 4x4 square gri

Question

16 numbers (not necessarily distinct) are filled in the 4x4 square grid. What is the sum of the four numbers in the corners of the grid?

1) Numbers that are filled in the grid are either 1 or 2.

2) The sum of the numbers in any 2x2 square (inside the 4x4 square grid) is 7.

sauravleo123 · Answer

Note that any numbers are allowed. They can be distinct or same.
Statement 1: all numbers can be 1
Sum of corners=4
All numbers can be 2
Sum of corners=8
Insufficient
Statement 2: Try 2,2,2,1
Sum of corners=7
Try 1,1,1,4
Sum of corners=7
Try 2,1,4,0
Sum of corners=7
Try 6,1,0,0
Sum of corners=7
Try 9,-2,0,0
Sum of corners=7
Try 7,0,0,0
Sum of corners=7
Sufficient
B

nick1816 · Answer

Statement 1-

Case 1
a+d+m+p=1+1+1+1=4

Case 2

a+d+m+p=2+2+2+2=8

Insufficient

Statement 2-

Sum of all 16 numbers = a+b+c+d.....+o+p= 4*7=28 .......(1)

Also f+j+g+k= g+k+h+l = 7

f+j= h+l ......(2)

Hence, e+i+h+l = 7

Similarly b+c+n+o=7 ......(3)

And f+g+j+k = 7 .......(4)

From (1), (2), (3) and (4)

a+d+m+p = 28-7-7-7 = 7

Sufficient

funkyakki · Answer

Hi. for the 2nd statement, how can we assume that there are only 4 2x2 grid squares in total?

nick1816 · Answer

funkyakki

I never assumed that there are only 4 2 by 2 squares in the grid. I considered those 4 so that i can get the sum of all the 16 numbers, as there is no overlapping in those squares and they cover all the 16 numbers.

BTW there are total 9 two by two squares in the grid.

Thekingmaker · Answer

this is a pretty easy question oncce you note down all the possiblities for example being
A - 1212 -6, 1112 - 5 , 2221-7 , 1111- 4 , 2222 - 8 , these are the many possiblities 
B however if the square in 2*2 matrix then it would be 2 or one in the corner and it will add upto 
IMO B i hope it helps

Crytiocanalyst · Answer

1) Numbers that are filled in the grid are either 1 or 2.
The possibilities could be 4 , 8, 7 , 5 
Clearly insufficient therefore out

2) The sum of the numbers in any 2x2 square (inside the 4x4 square grid) is 7.
Since the 2*2 matrix should be exactly 7 it should be filled with 1 and 2 always no other possibilities

and the 4 corners should be filled with 2,2,1,1 to satisfy the sum 
giving us 6 
Clearly sufficient

Therefore IMO A

Pritishd · Answer

I don't think that there are no other possibilities besides 1 and 2 as you have called out. We can also fill using integers 1, 1, 2 and 3 and still get a sum of 7 for any 2x2 square. Below is one of the arrangements
1, 2, 1, 2
1, 3, 1, 3
1, 2, 1, 2
1, 3, 1, 3

The sum of the numbers in the corner i.e. 1 + 2 + 3 + 1 = 7. We get the same result when we arrange 1 and 2
1, 2, 1, 2
2, 2, 2, 2
1, 2, 1, 2
2, 2, 2, 2
The sum of the numbers in the corner i.e. 1 + 2 + 2 + 2 = 7

Let me know your thoughts!

hadimadi · Answer

Hi,

(A) Fill the corners with only 2's in one case, and only 1's in the other, and we get a contradiction.

(B) We look for the value of a+d+m+p

Draw a 4x4 grid and name the squares a-p.

Taking a look only at the the first two rows of the grid, we know that:

a+b+e+f=b+c+f+g -> a+e=c+g, and equivalently, that b+f=d+h, and in total, that

a+e+d+h=7 (Eq.1)

We can replicate that procedure for the second and third row, and we get:

e+i+h+l=7 (Eq.2)

Doing the same for third and fourth row, we get:

i+m+l+p=7 (Eq.3)

Eq.1 + Eq.3 <-> (a+d+m+p)+(e+h+i+l)=14 (Eq.4)

Eq.4 - Eq.2 <-> a+d+m+p=7

So (B) is the correct answer

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16 numbers (not necessarily distinct) are filled in the 4x4 square gri

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