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(53)

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(53)

(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