Question

Put the numbers 1, 2, 3, 4, 5, 6, & 7 in the circles so that each straight line of three
numbers add up to the same total.

Answer 1

HELLO DEAR,


YOU CAN FIND IN THE GIVEN ATTACHED,

THE SUM OF EVERY NO. ON EACH LINE IS EQUAL (I.E, 12)

$\mathbf{2 + 4 + 6= 12.}$
$\mathbf{1 + 4 + 7 = 12.}$
$\mathbf{5 + 4 + 3 =12.}$




I HOPE ITS HELP YOU DEAR,
THANKS

Answer 2

Given : 1,2,3,4,5,6 and 7 numbers & circles

To find : Put the numbers so that the each straight line  of the three numbers add up  to the three numbers

Solution:

Let say A , B , C , D , E  , F  & G are 7 numbers  from 1 to 7

& D is in center

A & G , B & F , C & E are opposite

A + D + G  = B + D + F  = C + D  + E

=> A + G = B + F = C + E

Adding all numbers

A + B + C + D + E + F + G  + 2D

( A + B + C + D + E + F + G = 28  ( 1 to 7 sum)

= 28 + 2D

28 + 2D  is Divisible by  3  as sum of 3 Equal sums

Hence

D = 1   ,  4   ,   7  

Case 1  

D = 1

28 + 2D = 30    Sum of Each   = 10   ,  D = 1

=> A + G = B + F = C + E  = 9

only possible solution

2 + 7 = 3 + 6  = 4 + 5

Case 2  

D = 4

28 + 2D = 36    Sum of Each   = 12   ,  D = 4

=> A + G = B + F = C + E  = 98

only possible solution

21+ 7 = 2 + 6  = 3 + 5

Case 3

D = 7

28 + 2D = 42    Sum of Each   = 14   ,  D = 7

=> A + G = B + F = C + E  = 97

only possible solution

1 + 6 = 2 + 5  = 3 + 4

Hence 3 possible Solutions with center numbers  1 , 4 , 7

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