Question

Make every circle add up to 0 using given integers.
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Answer 1

Solution :-

→ First circle given :- (2 + 4 - 6) = 0

from first circle starting anti - clockwise now,

→ Second circle = (-6) + 3 + x = 0 => (-3) + x = 0 => x = 3 .

so,

→ Third circle = 3 + (-2) + x = 0 => 1 + x = 0 => x = (-1)

so,

→ Fourth circle = (-2) + x + y

we can take ,

• x and y = (7 , -5) or (6, -4)

if we take 7 , (-5) :-

→ (-2) + 7 + (-5) = 0

also, taking 7 in middle of circle :-

→ Fifth circle = (-5) + (-3) + x = 0 => (-8) + x = 0 => x = 8

→ Sixth circle = 8 + (-1) + x = 0 => 7 + x = 0 => x = (-7)

→ Seventh circle = (-7) + 2 + x = 0 => (-5) + x = 0 => x = 5 .

taking (-5) in middle of circle :-

→ Fifth circle = 7 + (-3) + x = 0 => 4 + x = 0 => x = (-4)

→ Sixth circle = (-4) + (-1) + x = 0 => (-5) + x = 0 => x = 5

→ Seventh circle = 5 + 2 + x = 0 => 7 + x = 0 => x = (-7) .

if we take (6, - 4) in fourth circle :-

taking 6 in middle of circle :-

→ Fifth circle = (-4) + (-3) + x = 0 => (-7) + x = 0 => x = 7

→ Sixth circle = 7 + (-1) + x = 0 => 6 + x = 0 => x = (-6)

→ Seventh circle = (-6) + 2 + x = 0 => (-4) + x = 0 => x = 4 .

taking (-4) in middle of circle :-

→ Fifth circle = 6 + (-3) + x = 0 => 3 + x = 0 => x = (-3)

→ Sixth circle = (-3) + (-1) + x = 0 => (-4) + x = 0 => x = 4

→ Seventh circle = 4 + 2 + x = 0 => 6 + x = 0 => x = (-6) .

[Note :- Given integers doesnt not have some required values. ]

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