Question

One-sixth of the smallest of three consecutive even

Answer 1

Answer:

The three even, consecutive integers are 72, 74, and 76.

Step-by-step explanation:

Let a be the first even integer.

Then the other two consecutive integers, b and c, will be represented by:

$\displaystyle b = a + 2 \text{ and } \\ \\ c = b + 2 = (a + 2) + 2 = a + 4$

One-sixth of the smallest (that is, a) is three less than one-tenth the sum of the other two even integers (that is, b and c).

Therefore:

$\displaystyle \frac{1}{6} a = \frac{1}{10}\left( b +c\right) - 3$

Solve for a. Substitute:

$\displaystyle \frac{1}{6} a =\frac{1}{10}\left((a+2)+(a+4)\right) - 3$

Simplify and solve for a:

$\displaystyle \begin{aligned} \frac{1}{6} a &= \frac{1}{10}(2a + 6) - 3 \\ \\ \frac{1}{6} a &= \frac{1}{5} a + \frac{3}{5} - 3\\ \\ -\frac{1}{30} a &= -\frac{12}{5} \\ \\ a &= 72\end{aligned}$

Hence, the first even integer is 72.

Therefore, the two other consecutive even integers must be 74, and 76.

In conclusion, the three even, consecutive integers are 72, 74, and 76.