You find the highest number that can divide into all of them without leaving any with a remainder.
Hope this helps.
7-4+<u>3x0</u>+1=<u>7-4</u>+0+1
=<u>3+0</u>+1
=<u>3+1
</u>=4<u>
</u>
Lets x is the 1st consecutive even integers
x +2 :the 2nd consecutive even integers
x +4: the 3rd consecutive even integers
So sum of all 3 equals -78 then
x + (x+2) + (x+4) = -78
x + x + 2 + x + 4 = -78
Simplify
3x + 6 = -78
Subtract (-6) to both sides
3x +6 -6 = -78 - 6
Simplify
3x = -84
Divided both sides by 3
3x/3 = -84/3
x = -28
x + 2 = -28 + 2 = -26
x +4 = - 28 + 4 = -24
Double check: -28 + -26 + -24 = -78
So the three consecutive even integers are: -24, -26, -28
The smallest integer would be -28
1 and 3, same explanation as last time
10 = 10^1
10 x 10 = 100 = 10^2
10 x 10 x 10 = 1000 = 10^3
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Therefore:
6,000 = 6 x 10^3