Find three consecutive integers such that 5 times the first is 5 more than the sum of the second and third integers.

Question

Find three consecutive integers such that 5 times the first is 5 more than the sum of the second and third integers.

Answer 1

There is no solution, which we can prove as follows.

The statement gives us the equation:

5x = 5 + (y+z)

The fact that the integers are consecutive gives us that

y = x+1

z = y+1 = x+2

Upon substitution,

5x = 2x+8, implying that x = 8/3, which is not an integer. Because we have three independent equations, this is the only solution. Hence, no solution exists where x,y, and z are integers.

Answer 2

The 3 consecutive integers are

n (first)

n + 1 (second)

n + 2 (third)

The wording of the problem translates into

5n = (n + 1) + (n + 2) + 5

5n = 2n + 8

3n = 8

This does not give an integer. Is this problem worded correctly?