Find the number of ordered pairs $$(a,b)$$ such that $$a, b\in\{10,11,\cdot\cdot\cdot,29,30\}$$ and $$\gcd(a,b)+\operatorname{lcm}(a,b)=a+b$$.

Let $$g=\gcd(a,b)$$. Write $$a=g\,x,\qquad b=g\,y,\qquad$$ where $$\gcd(x,y)=1$$.

The given condition becomes
$$g + gxy = gx + gy$$
Divide by $$g \;(\,g\gt 0\,)$$ to get
$$1 + xy = x + y$$
Rearrange:
$$xy - x - y + 1 = 0 \;\;\Longrightarrow\;\; (x-1)(y-1)=0$$

Hence $$x=1$$ or $$y=1$$.

Case $$x=1$$: $$a=g,\; b=gy$$. Case $$y=1$$: $$b=g,\; a=gx$$.

Thus one of the two numbers is an exact multiple of the other; equivalently, one divides the other. Conversely, if one number divides the other, then $$\gcd(a,b)=\text{smaller}$$ and $$\operatorname{lcm}(a,b)=\text{larger}$$, so the equality $$\gcd+\operatorname{lcm}=a+b$$ holds.

Therefore we must count all ordered pairs $$(a,b)$$ with $$10\le a,b\le 30$$ such that one divides the other.

List each integer in $$[10,30]$$ and the multiples of that integer which still lie in $$[10,30]$$:

10 → 20, 30
11 → 22
12 → 24
13 → 26
14 → 28
15 → 30
16 - 30 → no new multiples within the range

Unordered distinct pairs obtained are
$$(10,20),(10,30),(11,22),(12,24),(13,26),(14,28),(15,30)$$ — a total of $$7$$ pairs.

Each such pair contributes $$2$$ ordered pairs (e.g. $$(10,20)$$ and $$(20,10)$$), so distinct pairs give $$7\times 2 = 14$$ ordered pairs.

Finally, the pairs with equal entries $$(a,a)$$ also satisfy the condition (because $$\gcd=\operatorname{lcm}=a$$). There are $$30-10+1 = 21$$ such pairs.

Total ordered pairs $$=14+21=35$$.

Hence the required number of ordered pairs is 35.