Algebra Readiness — Variables, Substitution & Cryptarithms

By the end of this chapter you will:

1. Translate a word problem into a one-variable equation in under a minute
2. Solve linear equations using the balance method, fluently
3. Crack cryptarithms (letter-digit puzzles) by column-by-column reasoning
4. Use bounded search to find all integer solutions

4.0 Why this chapter is the biggest leverage point

BC baseline: G5 teaches one-step equations with one variable like $4 + X = 15$. G6 reinforces this with the balance method. But BC stops there — there are no two-step equations (e.g. $2(x+3) = 5x - 9$), no cryptarithms, and no bounded enumeration. Gauss expects all three.

Every recent paper has 3–4 questions that are trivial with algebra and slow without.

2024 Q2 starts with the gentlest possible: "If $n = 5$, the value of $n + 2$ is …" — just substitution. By Q12 the same skill solves "The length of a rectangle is twice its width. The perimeter is 120 cm. The width is …" in two lines.

🔑 The mental shift: stop asking "what number is it?" Start writing "let $x$ = the thing I don't know" — then the problem hands you the equation.

4.1 From words to a variable — the translation table

⚠️ Trap: "Five less than $x$" means $x - 5$, not $5 - x$. Read direction carefully.

Substitution warm-up — 2024 Q2

"If $n = 5$, the value of $n + 2$ is…"

Substitute: $n + 2 = 5 + 2 = \mathbf{7}$. (C). ✅

A six-second problem if you don't overthink it. Notice: Gauss starts almost every paper with a substitution Q to verify the student can do this.

4.2 The balance method — solving one-variable equations

Rule: whatever you do to one side, do to the other. The equation stays "balanced".

Example — solve $3x + 7 = 22$

$$3x + 7 = 22 \implies 3x = 15 \implies x = 5$$

Example — solve $5x - 4 = 2x + 11$

Subtract $2x$ from both sides: $3x - 4 = 11$.
Add $4$ to both sides: $3x = 15$.
Divide by $3$: $x = 5$.

4.3 Setting up from a word problem — 2024 Q12

"The length of a rectangle is twice its width. The perimeter of the rectangle is $120$ cm. The width of the rectangle is …"

Step 1 — Name the unknown. Let $w$ = the width (in cm).

Step 2 — Express what you know in terms of it.

• Length $= 2w$
• Perimeter $= 2(\text{length}) + 2(\text{width}) = 2(2w) + 2w = 6w$

Step 3 — Write the equation from the constraint.

$$6w = 120$$

Step 4 — Solve.

$$w = 20 \text{ cm}$$

Answer (A). ✅

🔑 Always name the variable first. Half the difficulty of word problems disappears once you pin down what $x$ stands for.

4.4 Inverse mean — 2024 Q13

"Eloise purchased a number of water hand pumps at a mean price of $85 per pump. She spent a total of $765. How many pumps did she purchase?"

Recall: $\text{mean} = \dfrac{\text{total}}{\text{count}}$, so $\text{count} = \dfrac{\text{total}}{\text{mean}}$.

Let $n$ = count of pumps.

$$\frac{765}{n} = 85 \implies n = \frac{765}{85} = 9$$

Answer (C). ✅

🔑 Inverse mean: when the mean is given and you want the count, divide total by mean. When the mean is given and a single missing item is asked, multiply mean by count then subtract the known sum.

4.5 Cryptarithms — letters stand for digits

A cryptarithm is an arithmetic puzzle where each letter is a single digit, and the same letter means the same digit throughout. The trick is column-by-column reasoning with carries / borrows.

2024 Q11 — worked in full

In the subtraction of two-digit numbers, the letters $P$ and $Q$ each represent a single digit:

$$\begin{array}{r} 8\,P \\ -\;Q\,6 \\ \hline 4\,9 \end{array}$$

Find $P + Q$.

Step 1 — Units column. We need $P - 6 = 9$ (units digit). That requires $P = 15$, which is impossible for a single digit. So we must have borrowed from the tens column: $P + 10 - 6 = 9 \implies P = 5$.

Step 2 — Tens column. After lending $1$ to the units column, the tens row has $8 - 1 = 7$ available. So $7 - Q = 4 \implies Q = 3$.

Step 3 — Verify. $85 - 36 = 49$. ✓

$$P + Q = 5 + 3 = \mathbf{8}$$

Answer (D). ✅

Cryptarithm playbook

1. Start at the units column. It has no carry coming in from the right.
2. Track each carry / borrow explicitly.
3. Use bounds: each letter is a digit $0$–$9$, and leading letters are $1$–$9$.
4. Verify by plugging back in. Always.

4.6 Bounded search — when algebra alone isn't enough

Sometimes the equation has multiple integer unknowns and you can't isolate one. Then enumerate the smaller side.

Example

Find all positive integers $a, b$ with $3a + 5b = 50$.

$a$ must be small enough that $5b$ stays positive: $a \le 16$.

Also, $5b = 50 - 3a$ must be divisible by $5$, so $3a$ must be divisible by $5$, so $a$ must be a multiple of $5$: $a \in \{5, 10, 15\}$.

Three solutions: $(5, 7), (10, 4), (15, 1)$.

🔑 Bounded-search technique: combine divisibility (from coefficients) with positivity bounds. Often you go from "infinite solutions" to "a 3-row table" in two lines.

4.7 Trap Alerts ⚠️

1. "Less than" reverses the order. "5 less than $x$" is $x - 5$, not $5 - x$.
2. Don't forget to define your variable. Writing "$x = 20$" without saying "let $x$ = width" loses marks on Olympiad solutions and confuses you mid-problem.
3. Carries and borrows are easy to miss. In Q11 above, students who forget the borrow get $P = -1$, which is impossible, and conclude "no answer" instead of going back.
4. Parity in mixed problems — "$3a + 5b = 50$": since $50$ is even and $3a, 5b$ have parity tied to $a, b$, you can also reason mod 2 to cut cases.

4.8 Mnemonic

"Name it, express it, equate it, solve it."

• Name the unknown (let $x$ = …)
• Express the other quantities in terms of $x$
• Equate using the constraint sentence
• Solve by balance method

Practice Set (10 problems, mixed difficulty)

1. (Part A) If $a = 7$, find $3a - 4$.
2. (Part A) Solve: $4x - 3 = 17$.
3. (Part A) Sarah is $4$ years older than her brother. The sum of their ages is $24$. How old is Sarah?
4. (Part B) Solve: $2(x + 3) = 5x - 9$.
5. (Part B) A rectangle has length $3$ cm more than its width. Its perimeter is $54$ cm. Find the width.
6. (Part B) In the cryptarithm $\overline{AB} + \overline{BA} = 121$, find $A + B$.
7. (Part B) The sum of three consecutive integers is $96$. What is the largest?
8. (Part C) Find all positive integer solutions to $2x + 7y = 50$.
9. (Part C) In $\overline{ABC} \times 6 = \overline{DDDD}$, where $A, B, C, D$ are digits, find $A + B + C + D$.
10. (Part C) Three numbers have mean $20$. The smallest is $12$ and the largest is twice the middle one. Find the middle number.

Answers: 1) 17; 2) 5; 3) 14; 4) 5; 5) 12; 6) 11; 7) 33; 8) $(1,\!\frac{48}{7})$ not integer, valid: $(8,2), (1,?),$ — check: $7y$ ≤ 50 → $y$ ∈ {2,4,6}: $(18,2)?$ — re-derive: $7y \le 50 - 2 \cdot 1 = 48$. $y=2: 2x=36, x=18$; $y=4: 2x=22, x=11$; $y=6: 2x=8, x=4$. So $\{(18,2),(11,4),(4,6)\}$; 9) Try $\overline{DDDD}/6$: $1111/6$ not integer, $2222/6$ not, $3333/6$ not, $4444/6$ not, $5555/6$ not, $6666/6 = 1111$, so $A=1,B=1,C=1,D=6$: sum $= 9$. $\frac{7777}{6}$ not, $\frac{8888}{6}$ not, $\frac{9999}{6}$ not. ✓; 10) Let middle = $m$, largest = $2m$. Sum: $12 + m + 2m = 60 \Rightarrow m = 16$.

End of chapter. Next: Geometry I — Angles, Triangles & Symmetry.