Algebra & Patterns — Turn the story into a letter.

Here's a sentence: Tom has $40 more than Jerry, and together they have $120. Most kids stare at it, then start guessing pairs of numbers. You don't have to. The story is hiding one equation, and the second you give the unknown a name, the equation falls out.

Watch what each phrase becomes. Algebra word problems are word problems with one extra step — name the unknown. Once you write “let \(x\) = the number of apples,” every phrase becomes a fact about \(x\):

• “twice as many as” → \(2x\)
• “5 more than” → \(x + 5\)
• “5 less than” → \(x - 5\) (NOT \(5 - x\) — ‘less than’ flips the order)
• “5 less than twice \(n\)” → \(2n - 5\) (build ‘twice \(n\)’ first, then take 5 off)
• “one less than half of” → \(\tfrac{x}{2} - 1\)
• “three times the sum of \(x\) and 7” → \(3(x + 7)\)
• “the difference between A and B” → \(A - B\) (order matters)
• “product” → multiply; “quotient” → divide

Drill this until it's automatic. The trick is having the courage to write the equation before you know how to solve it.

Draw the bar — let the picture write the equation

The bar model works whenever one amount is described in terms of another. Draw a bar for the unknown, then build every other quantity out of that same bar.

Tom has $40 more than Jerry; together $120.

Cover the gold $40 piece with your finger. What's left is two equal bars summing to \(120 - 40 = 80\). So each bar = $40 (Jerry), and Tom = \(40 + 40 = 80\). The picture showed the equation: \(2(\text{Jerry}) + 40 = 120\).

One is a multiple of the other. A has three times as much as B; together $200.

B is one bar, A is three. That's four equal bars adding to $200, so one bar = \(200 \div 4 = 50\). B has $50; A has \(3 \times 50 = 150\).

A chain of three. Anna has twice as many stickers as Ben; Ben has 5 more than Carla; together 55.

Carla is one bar. Ben is that bar plus 5. Anna is two bars plus 10. Stack them: 4 bars plus 15 makes 55, so 4 bars = 40 and one bar = 10. Carla 10, Ben 15, Anna 30.

Build every quantity out of the same bar, then the total hands you the bar.

Now SOLVE it — an equation is a balance scale

You named the unknown and wrote the equation. To finish, you have to get the letter alone. Here's the one idea behind every step you'll ever take: an equals sign is the bar of a balance scale. The two sides weigh exactly the same. So long as you do the same thing to both pans, it stays level.

Want \(x\) by itself? On the left it's sitting next to a weight of 3. Lift that 3 off — but the scale only stays level if you lift 3 off the right pan too. Left becomes \(x\); right becomes \(8 - 3 = 5\). That's the whole reason “subtract 3 from both sides” is legal: you're keeping the scale balanced.

Two weights on a pan, solved by halving. For \(3x = 12\), the left pan holds three equal \(x\)-blocks balancing 12. Split both pans into 3 equal shares: one \(x\)-block balances \(12 \div 3 = 4\). Dividing both sides by 3 is the same as sharing each pan into 3 fair piles.

Match-and-cancel: weights on BOTH pans. Sometimes the same thing sits on both pans already. If \(x + 5 = 2x + 1\), there's an \(x\) on each side — lift one \(x\) off both. Left becomes 5, right becomes \(x + 1\); then lift 1 off both to get \(x = 4\). Cancel what's common to both pans first; it shrinks the problem before you even start.

Watch it on a real contest one (Kangaroo 2010, problem 1). The puzzle balances \(\triangle + \triangle + 6\) against \(\triangle + \triangle + \triangle + \triangle\). Two triangles sit on both pans — lift them off both. What's left is \(6\) balancing \(\triangle + \triangle\). Split both pans in two: one triangle weighs \(6 \div 2 = \textbf{3}\). Cancelling the common weight did all the work.

Pictured-intuition adapted from Competition Math for Middle School (AoPS).

Two things stacked on the letter — peel from the OUTSIDE in

Most contest equations aren't one step. Look at \(8t + 9 = 65\): the letter \(t\) is wearing two layers — it got multiplied by 8, and then a 9 was added on top. To free it, take the layers off in the reverse order you'd put them on: the +9 went on last, so it comes off first. Same “socks then shoes” logic you'll meet again in Chapter 7.

Tidy each side BEFORE you peel

Sometimes a side is messy before you even start: \(2x + 3 - x + 5 = 14\). Don't touch the balance yet — first collect like terms on that side, the way you'd gather all the apples into one pile. The \(2x\) and \(-x\) are the same object (\(x\)), so they merge to \(x\); the 3 and 5 merge to 8. The whole left side is just \(x + 8\). Now the equation reads \(x + 8 = 14\), and one peel finishes it: \(x = 6\).

You can only combine terms that are the same kind: \(2x\) and \(5x\) merge to \(7x\), but \(2x\) and \(5\) never do — one counts \(x\)'s, the other is a plain number. Simplify first, and a scary-looking equation usually shrinks to one you've already beaten.

Combine-like-terms and peel-order framing inspired by AoPS Prealgebra.

You may not need the letter at all

Here's a move that feels like cheating. Suppose \(3x - 2 = 11\) and the question wants \(6x + 5\). Your reflex is to solve for \(x\) first. You don't have to. Peel one layer: \(3x = 13\). The thing you want is built from \(6x\), which is just double \(3x\) — so \(6x = 26\), and \(6x + 5 = \textbf{31}\). You never found \(x\) (it isn't even a whole number), and you didn't need to.

Find-the-expression-not-the-variable framing inspired by AoPS Prealgebra.

You open a problem and there's a symbol you've never seen — \(\star\), \(\diamond\), \(\spadesuit\), \(\triangle\) — with a definition next to it. Your stomach drops. Don't let it. That symbol is a machine: it eats two numbers and spits out one, using the recipe the problem handed you.

Put the LEFT number on the left of the recipe, the RIGHT number on the right, then compute. That's the whole game.

Order can matter

Try three recipes on the same two inputs and watch what happens when you swap them:

See the pattern? When the recipe treats the two inputs the same way, swapping does nothing. When one input gets squared or doubled and the other doesn't, swapping changes the answer. So read which number is on the left.

Nested operations — work inside out

Look at 1, 4, 7, 10, 13. What's the 100th number? You could write out all 100 — or you could notice the only thing that's happening: +3, every step. Once you see the step, you can jump straight to any term.

From the 1st term to the 5th, the step is taken 4 times, not 5.

The off-by-one trap. The first term hasn't moved from itself yet, so it's \(a_1 + 0\cdot d\). The 100th is 99 steps out. Always test your formula on \(n = 1\): it should spit back the real first term.

Walkthrough. Find the 50th term of 7, 11, 15, 19, …

• Start \(a_1 = 7\), step \(d = 4\).
• \(a_{50} = 7 + (50-1)\times 4 = 7 + 196 = \textbf{203}\).
• Test: \(a_1 = 7 + 0\cdot 4 = 7\). ✓

See the step as dots — growing figures

The same “same step every time” idea shows up as a picture that grows. Look at a row of dot-triangles, each one adding a new bottom row:

Each figure is the one before it plus a fresh bottom row. The new row is what you count.

You don't redraw the whole picture. You ask: what does the new ring or row add? Here the rows added are 2, 3, 4, … (these are the triangular numbers 1, 3, 6, 10). For shapes that grow by a border, the border is a small arithmetic sequence of its own — count only the new piece, then add it to the last total.

Watch it on a real one (AMC 8 2020, problem 4). Hexagons grow by one band of dots each time. A hexagon has 6 sides, so each new band holds 6 more dots than the last: the bands run 1, 6, 12, 18, … The third hexagon already shows \(1 + 6 + 12 = 19\) dots. The next one only tacks on the band of 18: \(19 + 18 = \textbf{37}\). You never drew the fourth hexagon — you only counted its outer ring.

When there's no fixed step — shrink it, table it, read the jumps

Not every pattern marches by the same step. Some sequences speed up. The move that cracks all of them has a name: Reduce and Expand. When a problem throws a scary number at you — “the 100th figure,” “a 26-team league” — you don't fight the big number. You shrink it to the first few cases you can actually count, line them up in a table, and watch what happens. The pattern hides in the jumps between terms, not the terms themselves.

Take the triangular numbers — dots stacked in a growing triangle: 1, 3, 6, 10, 15. They don't grow by a fixed amount. So look one level down, at the differences:

The terms 1, 3, 6, 10, 15 look hard to predict, but the jumps are dead simple: 2, 3, 4, 5, … — the counting numbers. To get the 6th term you don't need a formula; the next jump is \(+6\), so the 6th triangular number is \(15 + 6 = 21\). That's the whole Reduce-and-Expand idea: when the terms won't talk, ask the differences.

Strategy framing inspired by Posamentier, The Art of Problem Solving (teacher resource).

Two equations, two unknowns. It looks like twice the work — but the whole game is to get rid of one unknown so you're back to a single equation you already know how to crush. There are two ways to do it, and one read of the problem usually tells you which.

The sum-and-difference shortcut

When the system is \(x + y = S\) and \(x - y = D\):

• Add: \(2x = S + D\), so \(x = \dfrac{S+D}{2}\) — the average.
• Subtract: \(2y = S - D\), so \(y = \dfrac{S-D}{2}\) — the half-difference.

Contests use sum-and-difference constantly. Memorize it.

Don't always solve for everything

Your instinct is to find \(x\), then \(y\). Often you don't have to.

1. Coefficients swapped — add them. Find \(x + y\): given \(2x + 3y = 13\) and \(3x + 2y = 12\). Add them: \(5x + 5y = 25\), so \(x + y = 5\). You never found \(x\) or \(y\) alone — and you didn't need to.

2. Subtract them. Find \(x - y\), given \(3x + 2y = 20\) and \(x + 4y = 18\). Subtract: \(2x - 2y = 2\), so \(x - y = 1\).

3. Sometimes you nudge first. Given \(4x + y = 17\) and \(2x + 3y = 11\), adding cancels nothing. Double the second to \(4x + 6y = 22\), subtract the first → \(5y = 5\), so \(y = 1\), \(x = 4\), and \(x + y = 5\). Choosing the multiplier that kills a variable was the shortcut.

The factoring trick: when a product sneaks in

Some systems hide a factorisation. With whole numbers and \(x + y + xy = 14\): your instinct is to guess pairs. Resist it. Add 1 to both sides — \(x + y + xy + 1 = 15\) — and the left factors: \((x+1)(y+1) = 15\). The only factor pair above 1 is \(3\times 5\), so \(x\) and \(y\) are 2 and 4, giving \(x + y = 6\). The ‘+1’ on each variable is the whole trick.

When you see \(x + y + xy\), add 1 and factor.

(Factoring trick adapted from Problem Solving via the AMC, Australian Maths Trust.)

“Five positive numbers add to 50.” That single sentence is a sum constraint, and it's secretly a tug-of-war: the total is fixed, so the numbers can't all grow. If one goes up, another must come down.

Picture the total as one fixed pizza cut into 5 slices. Make one slice bigger and every other slice has to shrink — the pizza never gets larger.

The “distinct” trap

If the problem says distinct positive integers, you can't use four 1's. The smallest four distinct positive integers are 1, 2, 3, 4 — sum 10. So the largest would be \(20 - 10 = 10\). Big drop. Always re-read for the word “distinct.”

“What's the 2017th digit of this sequence?” The problem is daring you to write out 2017 terms. Don't take the bait. Sequences built by a fixed rule almost always loop — they fall into a short repeating block. Find the block, and 2017 becomes a tiny remainder problem.

Bend the line into a circle

Here's the picture that makes the remainder click. The terms don't run off forever in a straight line — once they loop, you can bend them into a wheel. Walk one step per term around the wheel; after a full lap you're back where you started, so step number \(n\) lands on the same spoke as its leftover after dividing by the cycle length.

The big number 98 never makes you take 98 steps. You take 94 steps from the first looped term, and 94 laps-worth of stepping is the same as its leftover, 4. Land on the spoke; read the term.

Warm up on something you already know: days of the week. Today is Monday — what day is it in 100 days? The week is a cycle of length 7. \(100 \div 7\) leaves a remainder of 2 (since \(98\) is a whole number of weeks), so you step 2 days past Monday: Wednesday. You did not count 100 days; you counted the leftover. That is the entire idea, dressed up.

How to find the cycle:

1. Compute the first several terms (4 to 6 is usually enough).
2. Watch for the moment the sequence (or a key feature) starts repeating.
3. The cycle length is the number of terms between two consecutive repeats.

Then reduce the position-number by the cycle length (its leftover) to land in the right spot.

Cycles you'll meet on contests:

• Units digits of powers (covered in the Number Theory lesson).
• Days of the week (cycle 7).
• Repeating decimals (e.g. \(1/7 = 0.\overline{142857}\), cycle 6).
• Folds or rotations that return to start.
• “Apply a rule over and over” — what does it return to?

“Think of a number. Double it. Add 5. Divide by 3. You got 7. What was the number?” Trying to guess-and-check is slow. There's a cleaner way: run the whole thing in reverse, undoing one step at a time.

Think of getting dressed. Socks, then shoes. To get back to bare feet, you take off shoes first, then socks. Last on, first off.

Quick: what's \(99^2 - 98^2\)? Squaring both feels brutal. But there's a back door. \(99^2 - 98^2 = (99+98)(99-98) = 197 \times 1 = 197\). The nasty subtraction collapsed into one multiplication. That back door has a name — and the cleanest way to see why all these tricks work is to draw them as areas.

First, why distributing works — it's an area

How do you multiply \(7 \times 103\) in your head? You split it: \(7\times 100 + 7\times 3 = 721\). That split is the distributive rule \(a(b+c) = ab + ac\), and a rectangle shows exactly why. Make a rectangle 7 tall and \(100 + 3\) wide. Slice it at the 100 mark: two rectangles, areas \(7\times 100\) and \(7\times 3\). Same total — you counted it in two pieces.

Whenever you “distribute,” you are slicing one rectangle into pieces and adding the pieces back. The rule never lies because area never changes when you cut it up.

Two binomials — the 2×2 box (this is FOIL)

Now multiply \((a+b)(c+d)\). Build a rectangle that's \(a+b\) tall and \(c+d\) wide. The two cuts — one across, one down — chop it into four pieces. Add the four areas and you have the product. No “FOIL” mnemonic to memorize; the grid shows all four products.

Try it with numbers: \((10+2)(10+3) = 12\times 13\). The four boxes are \(100, 30, 20, 6\), which add to \(156\). Check: \(12\times 13 = 156\). ✓ The box model and the multiplication agree because they're the same area, sorted two ways.

Why \(a^2 - b^2 = (a+b)(a-b)\) — cut and rearrange

Here's the picture that makes it obvious. Start with a big square, \(a\) on a side, so its area is \(a^2\). Bite a small \(b\times b\) square out of one corner. What's left — the gray L-shape — has area \(a^2 - b^2\).

Now cut the L into its two strips and lay them end to end. They form a single rectangle whose height is \(a - b\) and whose width is \(a + b\) — area \((a+b)(a-b)\). Nothing was added or thrown away, so \(a^2 - b^2 = (a+b)(a-b)\). The identity is the same gray region, re-laid as a rectangle.

Pictured-intuition adapted from Competition Math for Middle School (AoPS).

Watch it work:

• \(99^2 - 98^2 = (99+98)(99-98) = 197 \times 1 = \textbf{197}\).
• \(51^2 - 49^2 = (51+49)(51-49) = 100 \times 2 = \textbf{200}\).
• \(1000^2 - 999^2 = 1999\).

The reverse is just as useful: when you see a product like \((\text{big})(\text{small difference})\), it may be a hidden difference of squares.

Sum of consecutive odd numbers has the same flavor: \(1 + 3 + 5 + \cdots + (2n-1) = n^2\). Each new odd number wraps an L-shaped border around an \((n-1)^2\) square to grow it to \(n^2\).

Know the sum and product? You're done — without finding the numbers

Two secret numbers. You're told their sum and their product. The question asks for something else entirely — their squares added, their reciprocals added, their difference. Your instinct is to dig out the two numbers first. Resist it. The same square-of-a-sum identity, read backward, hands you the answer without ever knowing the numbers.

Start from the identity above: \((a+b)^2 = a^2 + 2ab + b^2\). Slide the \(2ab\) across and look what's left:

Watch it work. Two positive numbers have sum 5 and product 6. Find \(a^2 + b^2\). Plug in: \(5^2 - 2\cdot 6 = 25 - 12 = 13\). (Check: the numbers are 2 and 3, and \(4 + 9 = 13\). The identity beat you to it.) Want \(\tfrac{1}{a}+\tfrac{1}{b}\)? It's \(\tfrac{5}{6}\), no digging required.

Framing inspired by Competition Math for Middle School (AoPS).

Add up \(8 + 9 + 10 + 11 + 12\). You could grind it left to right — or notice the 10 sits dead center, and the rest pair up around it: \(8+12 = 20\) and \(9+11 = 20\), both twice the middle. The sum is then \(5 \times 10 = 50\). That one fact collapses most consecutive-integer problems.

Why it works — pair from the ends

Pair smallest with largest, 2nd-smallest with 2nd-largest, and so on. Each pair sums to the same thing — twice the middle. The middle pairs with itself; it's the fulcrum.

What if \(n\) is even? The middle is a half-integer.

For 4 consecutive integers 7, 8, 9, 10: the middle sits at 8.5, so sum \(= 4 \times 8.5 = 34\). Check: \(7+8+9+10 = 34\). ✓

For 6 consecutive integers summing to 2013: middle \(= 2013/6 = 335.5\). So the 3rd term is 335, the 4th is 336, and the integers are 333, 334, 335, 336, 337, 338. Largest \(= 338\).

Other consecutive-integer facts

Page-number problems — a special flavor

“How many digits to number pages 1 through 200?” Break into digit-length blocks:

For pages 1–200: \(9 + 180 + 3\times(200-99) = 9 + 180 + 303 = \textbf{492}\) digits.

Two consecutive numbers multiplied \(= X\)

If \(n(n+1) = 1056\), take \(\sqrt{1056} \approx 32.5\). Try \(32 \times 33 = 1056\). ✓ The integer just under \(\sqrt{X}\) is your \(n\).

The sequences so far had clean rules — add \(d\), multiply by \(r\), repeat a block. But contests also love sequences where each term is built from several earlier ones. The famous one: 1, 1, 2, 3, 5, 8, 13, … where each number is the sum of the two before it. You don't need a formula for these. You need a table.

Walkthrough. A sequence starts 1, 2 and each later term is the product of the previous two. Find the 6th term.

• \(a_1 = 1,\ a_2 = 2\).
• \(a_3 = 1 \cdot 2 = 2\).
• \(a_4 = 2 \cdot 2 = 4\).
• \(a_5 = 2 \cdot 4 = 8\).
• \(a_6 = 4 \cdot 8 = \textbf{32}\).

Function-iteration: apply the same rule over and over

A close cousin is function-iteration: start with a number, apply a rule, then apply it again to the result. The rule “double, then add 1,” starting at 3:

• \(2\cdot 3 + 1 = 7\)
• \(2\cdot 7 + 1 = 15\)
• \(2\cdot 15 + 1 = 31\)
• \(2\cdot 31 + 1 = 63\)

That's the recursive rule \(a_n = 2a_{n-1} + 1\) in disguise. When a problem asks for the rule applied many times, compute the first few and look for a cycle (Chapter 6).

The Fibonacci-stairs story

You're climbing a staircase, taking 1 stair or 2 at a time. How many ways to climb \(n\) stairs?

Why does \(f(n) = f(n-1) + f(n-2)\)? Look at the last step of any path to stair \(n\). It was either a 1-step (so you came from stair \(n-1\)) or a 2-step (from stair \(n-2\)). Every path falls into exactly one camp, so total ways \(= f(n-1) + f(n-2)\). This shows up in any “take 1 or 2 steps” counting problem — spot it once, recognize it forever.