Look for a Pattern

The ones digits run $8,6,4,2,0$ and then start over, a loop of length $5$. Since $43=8\times 5+3$, the $43$rd multiple sits at the third spot of a fresh loop, so it ends in $\boxed{{\displaystyle 4}}$.

Checking directly, $43\times 8=344$.

Dividing, $500=55\times 9+5$, so the hallway holds $55$ complete runs and then $5$ extra tiles.

Each complete run contributes one star, giving $55$ so far. The five leftover tiles are the first five of a new run, all hexagons, so none of them is a star.

$$55+0=\boxed{{\displaystyle 55}}$$

The ones digits run $3,9,7,1$, and then $3^5=243$ brings the digit back to $3$, and the loop restarts.

The loop is $3,9,7,1$, which is $\boxed{{\displaystyle 4}}$ steps long.

Since $43=10\times 4+3$, the exponent $43$ lands three steps into a fresh loop of $3,9,7,1$, and the third digit of that loop is $\boxed{{\displaystyle 7}}$.

The exponent could be in the millions and the work would not change, one division by the loop length.

Tracking only ones digits, $7\times 7$ ends in $9$, then $9\times 7$ ends in $3$, then $3\times 7$ ends in $1$, and the loop $7,9,3,1$ restarts.

Since $7=1\times 4+3$, the seventh power lands on the loop's third digit.

$$7^7\text{ ends in }\boxed{{\displaystyle 3}}$$

Indeed $7^7=823,543$.

Each terrace adds $3$, and getting from terrace $1$ to terrace $20$ takes $19$ such steps.

$$14+19\times 3=14+57=\boxed{{\displaystyle 71}}$$

In general the $n$th terrace holds the starting $14$ plus $n-1$ copies of the step.

Pair the top terrace with the bottom one, the second with the second from the bottom, and so on. Every pair totals the same amount,

$$14+71=17+68=\cdots =85$$

because each step inward adds $3$ to one partner and removes $3$ from the other. Ten pairs at $85$ apiece give

$$10\times 85=\boxed{{\displaystyle 850}}$$

Following the rule, $3\to 8\to 4\to 2\to 1\to 6\to 3$, and the display is trapped in a loop of six numbers, $3,8,4,2,1,6$.

With $100=16\times 6+4$, the $100$th number is the fourth member of the loop.

$$\boxed{{\displaystyle 2}}$$

Iteration problems almost always fall into a loop, because a rule with limited outputs must eventually revisit a number, and from there history repeats exactly.

The running totals are $1,4,9,16$, the perfect squares, so $n$ bursts should total $n^2$ beats. A fifth case agrees, $16+9=25$.

$$20^2=\boxed{{\displaystyle 400}}$$

Each new odd burst adds one more L-shaped shell to a square of dots, which grows it to the next square. The next insight draws it.

A shared flash needs a time that is a multiple of $6$ and of $10$ at once, and the smallest is $\text{lcm}(6,10)=30$. So the signs align every $30$ seconds, at $30,60,90,120,150,$ and $180$ seconds.

$$180÷30=\boxed{{\displaystyle 6}}$$

Two separate rhythms always merge into one bigger rhythm, and its length is their least common multiple, which you met back in the divisibility chapter.

Since $75=9\times 8+3$, nine full passes finish and the $75$th play is the third song of the next pass, song $\boxed{{\displaystyle 3}}$.

There are $33$ complete bands, each with $5$ blue rows, for $165$. The $4$ leftover rows are the start of a new band, and the first $5$ rows of any band are blue, so all $4$ leftovers are blue.

$$165+4=\boxed{{\displaystyle 169}}$$

Odd powers of $9$ end in $9$ and even powers end in $1$. Since $55$ is odd, $9^{55}$ ends in $\boxed{{\displaystyle 9}}$.

The ones digit alternates, $4$ for odd exponents and $6$ for even ones. The exponent $100$ is even, so the digit is $\boxed{{\displaystyle 6}}$.

Fourteen steps of $4$ sit between the first and fifteenth strands,

$$7+14\times 4=7+56=\boxed{{\displaystyle 63}}$$

The bottom row holds $5+29\times 2=63$ bricks. Pairing the top row with the bottom, the second with the second from the bottom, and so on gives $15$ pairs, each totaling $68$.

$$15\times 68=\boxed{{\displaystyle 1020}}$$

The screen cycles through $6,12,24,9,18,36,21$ and then repeats, a loop of length $7$.

Since $50=7\times 7+1$, the $50$th number is the first member of the loop, $\boxed{{\displaystyle 6}}$.

The running totals of odd numbers are the perfect squares, so the student stopped when $n^2=225$, that is $n=\boxed{{\displaystyle 15}}$.

The meeting point must be a multiple of both $8$ and $14$, and $\text{lcm}(8,14)=56$ teeth. The small gear completes

$$56÷8=\boxed{{\displaystyle 7}}$$

full turns. The big gear makes $4$, and the two marks line up again.

Both loops have length $4$, and $43$ leaves remainder $3$, landing on the third spot of each. So $3^{43}$ ends in $7$ and $7^{43}$ ends in $3$.

Then $7+3=10$, so the sum ends in $\boxed{{\displaystyle 0}}$.