Problema Solution
Find three consecutive integers whose product is 261 larger than the cube of the smallest integer.
Answer provided by our tutors
The three consecutive odd numbers can be written as 2k-1, 2k+3 and 2k+5 where k is integer.
(2k-1)*(2k+3)*(2k+5) = 261 + (2k-1)^3
the solution is not an integer thus there is no solution.
The three consecutive numbers can be written as x-1, x and x+1
(x-1)*x*(x+1) = 261 + x^3
x(x^2-1) = 261 +x^3
x^3 - x = 261 + x^3
x = - 261
The tree consecutive integer are -262, - 261, -260.