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

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Answer 1 · 2018-03-13T04:33:18

#61, 62, 63, 64#

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#

Answer 2 · 2018-03-13T04:37:30

61, 62, 63, 64

# (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#