Let $$a, b, c \in \mathbb{N}$$ and $$a < b < c$$. Let the mean, the mean deviation about the mean and the variance of the 5 observations $$9, 25, a, b, c$$ be $$18, 4$$ and $$\frac{136}{5}$$, respectively. Then $$2a + b - c$$ is equal to _____

We have five observations $$9, 25, a, b, c$$ with $$a < b < c$$ and $$a, b, c \in \mathbb{N}$$. Since the mean is 18,

$$\frac{9 + 25 + a + b + c}{5} = 18 \implies a + b + c = 56 \quad (i)$$

The mean deviation about the mean is 4, so

$$\frac{|9-18| + |25-18| + |a-18| + |b-18| + |c-18|}{5} = 4.$$

Evaluating the known terms gives

$$9 + 7 + |a-18| + |b-18| + |c-18| = 20 \implies |a-18| + |b-18| + |c-18| = 4 \quad (ii)$$

The variance is $$\tfrac{136}{5}$$, hence

$$\frac{(9-18)^2 + (25-18)^2 + (a-18)^2 + (b-18)^2 + (c-18)^2}{5} = \frac{136}{5}.$$

Since $$(9-18)^2 + (25-18)^2 = 81 + 49 = 130$$, it follows that

$$130 + (a-18)^2 + (b-18)^2 + (c-18)^2 = 136 \implies (a-18)^2 + (b-18)^2 + (c-18)^2 = 6 \quad (iii)$$

Let $$p = a-18,\; q = b-18,\; r = c-18$$. Then from (i), (ii), (iii) we get

$$p+q+r = 56 - 54 = 2,\quad |p|+|q|+|r| = 4,\quad p^2+q^2+r^2 = 6\,. $$

Because $$p+q+r=2$$ and $$|p|+|q|+|r|=4$$, the total negative contribution is $$\tfrac{4-2}{2}=1$$, so exactly one of $$p,q,r$$ equals $$-1$$ and the other two are positive summing to 3. Substituting one variable as $$-1$$ in the sum of squares gives

$$1 + q^2 + r^2 = 6 \implies q^2 + r^2 = 5,\quad q + r = 3\,. $$

Writing $$r = 3 - q$$ yields

$$q^2 + (3-q)^2 = 5 \implies 2q^2 - 6q + 9 = 5 \implies q^2 - 3q + 2 = 0 \implies q = 1 \text{ or } 2\,. $$

Thus $$(p,q,r) = (-1,1,2)$$ or $$(-1,2,1)$$. The ordering $a

$$a = 17,\quad b = 19,\quad c = 20\,. $$

Finally,

$$2a + b - c = 2(17) + 19 - 20 = 33\,. $$

The answer is $$\boxed{33}$$.