Question

write 8 consecutive integers whose sum is 4

Answer 1

Define x:

Let x be the smallest integer

The other 7 integers are (x + 1), (x + 2), (x + 3) , (x + 4) , (x + 5), (x + 6) and (x + 7)

Solve x:

Given that the sum is 4

x +  (x + 1) +  (x + 2) + (x + 3) + (x + 4) + (x + 5) + (x + 6) + (x + 7) = 4

x +  x + 1 +  x + 2 + x + 3 + x + 4 + x + 5 + x + 6 + x + 7 = 4

8x + 28 = 4

8x = 4 - 28

8x = -24

x = -3

Find the numbers:

Smallest number is -3

Therefore the other 7 numbers are -2, -1, 0, 1, 2, 3 and 4

Answer: The numbers are -3, 2, -1, 0, 1, 2, 3 and 4

Answer 2

Let x be the smallest integer. The other 7 integers are (x +1) + (x +2) + (x +3) + (x +4) + (x +5) + (x +6) and (x +7).

The sum is 4, x + (x +1) + (x +2) + (x +3) + (x +4) + (x +5) + (x +6) + (x +7) = 4, 8 x+ 28 = 4, 8x = 4-28= -24, Now x = -3,

the 8 consecutive integers for 4 is -3, 2,-1, 0,1,2,3 and 4.