How to Solve Math Problems by Finding a Pattern

Some problems are too big to attack head on. Nobody multiplies out the five hundredth term of anything. Instead you compute a few small cases, line the results up, and look for a pattern that repeats. Once you know how it repeats, one short division carries you to the five hundredth case.

Problem

Write out the ones digits of the first several multiples of $8$, that is $8, 16, 24, 32, 40, 48, \ldots$, and watch what the ones digits do. Then predict the ones digit of the $43$rd multiple of $8$.

Show a hint

The ones digits run $8, 6, 4, 2, 0$ and then start over, a loop of length $5$.
Divide $43$ by $5$ and let the remainder point to a spot inside the loop.

Show the full solution

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 $\boxed{4}$.

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

Problem

A mosaic artist lays tiles along a hallway in a repeating run of nine tiles, five hexagons, then three squares, then one star, over and over. Among the first $500$ tiles, how many are stars?

Show a hint

Each full run of $9$ tiles contains exactly one star, so first count complete runs inside $500$.
Since $500 = 55 \times 9 + 5$, there are $55$ full runs plus $5$ leftover tiles. Check whether any of those leftovers is a star.

Show the full solution

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.

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

Problem

Compute the ones digit of each of $3^1, 3^2, 3^3, 3^4, 3^5$, and keep going just far enough to see the ones digits repeat. How many steps long is the repeating loop?

Show a hint

The powers begin $3, 9, 27, 81, 243$. Read off just the ones digits.

Show the full solution

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 $\boxed{4}$ steps long.

Problem

Using the loop you just found, determine the ones digit of $3^{43}$.

Show a hint

The loop $3, 9, 7, 1$ restarts every $4$ powers, so divide $43$ by $4$.
A remainder of $3$ lands on the third digit of the loop.

Show the full solution

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 $\boxed{7}$.

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

Problem

The powers of $7$ have their own ones-digit loop. Find it, then find the ones digit of $7^7$.

Show a hint

Compute ones digits of $7^1, 7^2, 7^3, 7^4$. Multiplying ones digits only, $7, 9, 3, 1$, then the loop restarts.
With a loop of length $4$, the exponent $7$ leaves remainder $3$, the third spot.

Show the full solution

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.

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

Indeed $7^7 = 823{,}543$.

Problem

A vineyard climbs a hillside in terraces. The top terrace holds $14$ vines, and each terrace below holds $3$ more vines than the one above it. How many vines grow on the $20$th terrace from the top?

Show a hint

Make a tiny table. Terrace $1$ has $14$, terrace $2$ has $17$, terrace $3$ has $20$. Each step down adds one $3$.
Reaching terrace $20$ takes $19$ steps down from the top.

Show the full solution

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

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

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

Problem

How many vines grow on the hillside in total, across all $20$ terraces?

Show a hint

Pair the terraces from the outside in. The first with the last gives $14 + 71 = 85$, the second with the second to last gives $17 + 68 = 85$ again.
Twenty terraces form $10$ such pairs.

Show the full solution

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

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

Problem

A fidget counter shows $3$. Every button press changes the number by one rule. If the number is even, it is cut in half. If it is odd, $5$ is added. So the display reads $3, 8, 4, 2, \ldots$ as presses go by. What number shows after the $99$th press, that is, as the $100$th number in the list counting the starting $3$?

Show a hint

Keep pressing. The list runs $3, 8, 4, 2, 1, 6$ and then returns to $3$, so it loops every $6$ numbers.
Since $100 = 16 \times 6 + 4$, the $100$th number sits at the fourth spot of the loop.

Show the full solution

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.

$$\boxed{2}$$

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.

Problem

A drummer warms up in bursts. The first burst is $1$ beat, the second is $3$ beats, the third is $5$ beats, each burst two beats longer than the last. Compute the total beats after one burst, after two, after three, and after four, and stare at those totals. Then find the total number of beats after the $20$th burst.

Show a hint

The running totals are $1, 4, 9, 16, \ldots$, which are perfect squares.
After $n$ bursts the total is $n^2$, so evaluate $20^2$.

Show the full solution

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

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

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.

Problem

Two neon signs are switched on at the same moment. One flashes every $6$ seconds, the other every $10$ seconds. Over the next $3$ minutes after switch-on, how many more times will the two signs flash at exactly the same moment?

Show a hint

The joint flashes are themselves a repeating pattern. They happen at times that are multiples of both $6$ and $10$, and the first is at the least common multiple.
Shared flashes come every $30$ seconds, and $3$ minutes holds $180$ seconds.

Show the full solution

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.

$$180 \div 30 = \boxed{6}$$

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.

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Practice

A road-trip playlist of $8$ songs plays on repeat without shuffle. Counting from the first play, which song number is the $75$th song played?

Show the solution

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

Practice

A scarf is knitted in a repeating band of $12$ rows, $5$ blue rows followed by $7$ white rows. Among the first $400$ rows, how many are blue?

Show the solution

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.

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

Practice

What is the ones digit of $9^{55}$?

Show the solution

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

Practice

What is the ones digit of $4^{100}$?

Show the solution

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

Practice

A charm bracelet design puts $7$ charms on the first strand, and every following strand carries $4$ more charms than the one before. How many charms are on the $15$th strand?

Show the solution

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

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

Practice

A brick staircase has $30$ rows. The top row uses $5$ bricks, and each row below uses $2$ more bricks than the row above. How many bricks does the whole staircase use?

Show the solution

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 $68$.

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

Practice

A game screen starts at $6$ and updates with every tap. If the number on screen is less than $20$, it doubles. Otherwise, $15$ is subtracted. Counting the starting $6$ as the first number, what is the $50$th number to appear?

Show the solution

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, $\boxed{6}$.

Practice

Adding consecutive odd numbers starting from $1$, a student reaches a total of exactly $225$. How many odd numbers did the student add?

Show the solution

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

Practice

Two meshed gears have $8$ teeth and $14$ teeth, and a paint mark sits where they currently touch. As they spin, the marks meet again only after a whole number of teeth have passed on both gears. How many full turns does the small gear make before the two marks first line up again?

Show the solution

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

$$56 \div 8 = \boxed{7}$$

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

Practice

Using the ones-digit loops of $3$ and $7$, find the ones digit of $3^{43} + 7^{43}$.

Show the solution

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 $\boxed{0}$.

Look for a Pattern

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