$$3 \times 7^{22} + 2 \times 10^{22} - 44$$ when divided by 18 leaves the remainder _________.

We are asked to find the remainder obtained when the quantity $$3 \times 7^{22} + 2 \times 10^{22} - 44$$ is divided by $$18$$. In modular-arithmetic language, this means that we must evaluate the given expression modulo $$18$$, i.e. find its value in $$\pmod{18}$$.

First, we note that for any integers $$a$$ and $$b$$, the basic property of congruences is

$$a \equiv b \pmod{m} \;\;\Longrightarrow\;\; a \cdot c \equiv b \cdot c \pmod{m},$$

and

$$a \equiv b \pmod{m},\; c \equiv d \pmod{m} \;\;\Longrightarrow\;\; a \pm c \equiv b \pm d \pmod{m}.$$

We shall therefore replace each large power by its much smaller remainder modulo $$18$$ and then combine the results.

Step 1: Evaluate $$7^{22} \pmod{18}.$$

We first look for a pattern in the powers of $$7$$ modulo $$18$$:

$$7^1 = 7 \equiv 7 \pmod{18}.$$

$$7^2 = 49 \equiv 49 - 2\!\times\!18 = 49 - 36 = 13 \pmod{18}.$$

Now multiply again by $$7$$:

$$7^3 \;=\; 7^2 \times 7 = 13 \times 7 = 91 \equiv 91 - 5\!\times\!18 = 91 - 90 = 1 \pmod{18}.$$

We have reached $$7^3 \equiv 1 \pmod{18}.$$ This is extremely useful because now every third power will repeat the remainder $$1$$. In other words,

$$7^{3k} \equiv 1 \pmod{18}\quad\text{for every integer }k.$$

We write $$22$$ as a multiple of $$3$$ plus the remainder:

$$22 = 21 + 1 = 3 \times 7 + 1.$$

So

$$7^{22} = 7^{21 + 1} = 7^{21}\,7^1 = (7^3)^7 \, 7.$$

Because $$7^3 \equiv 1$$, we replace it by $$1$$:

$$(7^3)^7 \, 7 \equiv 1^7 \, 7 = 7 \pmod{18}.$$

Thus we have

$$7^{22} \equiv 7 \pmod{18}.$$

Step 2: Evaluate $$10^{22} \pmod{18}.$$

We again look for a pattern, this time in the powers of $$10$$ modulo $$18$$:

$$10^1 = 10 \equiv 10 \pmod{18}.$$

Now square that value:

$$10^2 = 100 \equiv 100 - 5\!\times\!18 = 100 - 90 = 10 \pmod{18}.$$

Notice that the remainder has come back to $$10$$. Therefore a second multiplication by $$10$$ will do the same:

$$10^3 = 10^2 \times 10 \equiv 10 \times 10 = 100 \equiv 10 \pmod{18}.$$

Hence once we reach the power $$2$$, every additional power leaves the remainder unchanged at $$10$$. Concretely, for all positive integers $$n \ge 1,$$

$$10^n \equiv 10 \pmod{18}.$$

In particular,

$$10^{22} \equiv 10 \pmod{18}.$$

Step 3: Substitute the reduced remainders into the original expression.

We have obtained

$$7^{22} \equiv 7 \pmod{18},$$

and

$$10^{22} \equiv 10 \pmod{18}.$$

Thus

$$3 \times 7^{22} + 2 \times 10^{22} - 44$$

becomes

$$3 \times 7 + 2 \times 10 - 44 \pmod{18}.$$

Carrying out the multiplications gives

$$21 + 20 - 44 \pmod{18}.$$

Add the first two numbers:

$$21 + 20 = 41,$$

so we now have

$$41 - 44 = -3 \pmod{18}.$$

Step 4: Convert the negative remainder to a positive one.

A remainder of $$-3$$ can be made positive by adding $$18$$ (because adding or subtracting any multiple of the modulus does not change a congruence):

$$-3 + 18 = 15.$$

Therefore

$$-3 \equiv 15 \pmod{18}.$$

The required non-negative remainder is thus $$15$$.

So, the answer is $$15$$.