How do you find four consecutive integers whose sum in twice the cube of 5?

2 Answers
Mar 13, 2018

61, 62, 63, 64

Explanation:

You can treat the four consecutive integers as (x), (x+1), (x+2), and (x+3).

You set up the equation:
(x) + (x+1) + (x+2) + (x+3) = 2 * (5^3)

Solve the equation:
4x + 6 = 250
4x = 244
x = 61

Now, you can plug in x back into your original four consecutive integers.

(61), (61+1), (61+2), (61+3),
which gives you:
61, 62, 63, 64

Mar 13, 2018

61, 62, 63, 64

Explanation:

(x)+(x+1)+(x+2)+(x+3)=2(5^3)

(x)+(x+1)+(x+2)+(x+3)=2(125)

(x)+(x+1)+(x+2)+(x+3)=250

So now we can just take out the parenthesis on the left side of the equation and simplify

(x)+(x+1)+(x+2)+(x+3)=250

4x+6=250

4x=244

x=61