STEP 1 OF 20 · Lesson Opening

Today: Simon's Favourite Factoring Trick (SFFT) for Unit Fractions

A 60-minute lesson that turns \(\frac{1}{x}\) + \(\frac{1}{y}\) = \(\frac{1}{n}\) into a clean factor-pair count.

📌 What you will learn today

Topic: Solving Diophantine equations of the form \(\frac{1}{x}\) + \(\frac{1}{y}\) = \(\frac{1}{n}\) (and variants) by Simon's Favourite Factoring Trick — an algebraic move that turns a sum into a product.
Category: Number Theory (NT) — sub-topic Diophantine · Unit Fractions.
Solves these AIMO problems: 2014 Q5 2008 Q3 2015 Q4 2006 Q1
Four past-paper problems — all unlocked by one move: add \(n^{2}\) to both sides.
AoPS Reference: Introduction to Number Theory by Mathew Crawford, Chapter 6 (Diophantine equations); Intermediate Algebra (SFFT chapters). The full unit-fraction technique is taught here from first principles.
Why this matters: Unit-fraction equations appear in 4+ AIMO papers across Q4–Q7 — the middle-difficulty band where 25-mark camp candidates earn or lose places. SFFT unlocks all of them with one recipe.
Time required: About 60 minutes for the full lesson, plus 30 minutes practising past papers afterwards.

How this lesson is structured

We will not start with formulas. We will start with a picture — pies cut into slices, and rectangles of integer area. Once you see why factoring works, the algebra writes itself.

Step 2 (Phase 0): Prerequisite knowledge check — unit fractions, common denominators, factor pairs, perfect-square differences.
Steps 3–5 (Phase 1): Visual intuition — pie slices, rectangles, the "magic \(n^{2}\)" geometric proof.
Step 6 (Phase 1.5): Formula manual — the SFFT recipe, with full derivation and generalisations.
Steps 7–9 (Phase 2): Three derivations — concrete (n=4), general n, and numerator > 1.
Steps 10–13 (Phase 3): Four worked examples in strict difficulty gradient (⭐ → ⭐⭐⭐⭐).
Step 14 (Phase 4): Three practice problems with hints and full solutions.
Steps 15–17 (Phase 5): Four real AIMO past papers in exam format.
Step 18 (Phase 5.5): Synthesis problem — combine SFFT with modular filtering.
Step 19: Mock test (3 problems, auto-graded).
Step 20: Cheat sheet + self-assessment.

Pedagogical note: SFFT is one of the most reused tricks in Olympiad number theory. Read each step at least twice. The arrow keys (← →) move between steps. Skip nothing in Phase 0 or Phase 1 — the picture you build now is what makes the algebra later feel inevitable.

STEP 2 OF 20 · Phase 0 · Prerequisite Knowledge

Phase 0 — Things you must already know

Quick self-check. If any of these feels shaky, pause here and patch the gap before continuing.

SFFT lives at the intersection of four small skills. None are hard alone, but together they do the heavy lifting of the lesson. Read each row, try the verification mentally, then continue.

Knowledge	Verification
0.1 Unit fractions — a unit fraction is \(\frac{1}{n}\) for some positive integer n.	You can read \(\frac{1}{7}\) as "one-seventh" and recognise that \(\frac{1}{m}\) + \(\frac{1}{k}\) is the sum of two such pieces.
0.2 Common denominators — combining fractions by finding a shared denominator.	\(\frac{1}{3}\) + \(\frac{1}{4}\) = \(\frac{4}{12}\) + \(\frac{3}{12}\) = \(\frac{7}{12}\).
0.3 Factor pairs — listing all ordered pairs (d, \(\frac{N}{d}\)) of positive integers whose product is N.	Factor pairs of 12: (1, 12), (2, 6), (3, 4), (4, 3), (6, 2), (12, 1). Six ordered pairs; three unordered.
0.4 Perfect-square difference — expanding products like (x − k)(y − k).	(x − 3)(y − 3) = xy − 3x − 3y + 9. Notice the +9 — that's a perfect square. This is the very identity SFFT exploits.

🔑 The setup of the trick is already in row 0.4. If you expand (x − 3)(y − 3) you naturally get the constant +9 = \(3^{2}\). SFFT works backwards: starting from xy − 3x − 3y, it asks "what constant should I add to make this factor cleanly?" The answer is always \(n^{2}\).

One quick warm-up

Try mentally: What are all the unordered factor pairs of 36?

(1, 36), (2, 18), (3, 12), (4, 9), (6, 6). Five unordered pairs.

And ordered? Multiply by 2 (each pair has two orderings) except (6, 6) which is its own mirror. So 5 × 2 − 1 = 9 ordered. Equivalently, the number of divisors of 36 is 9.

Hold on to this counting trick — it appears in every SFFT problem.

STEP 3 OF 20 · Phase 1 · Visual

Pie slices — what \(\frac{1}{x}\) + \(\frac{1}{y}\) = \(\frac{1}{n}\) really says

Forget letters. Start with two pies.

A unit fraction \(\frac{1}{n}\) is a single slice from a pie cut into n equal pieces. The equation \(\frac{1}{x}\) + \(\frac{1}{y}\) = \(\frac{1}{n}\) says: "two thinner slices, of sizes \(\frac{1}{x}\) and \(\frac{1}{y}\), together equal one thicker slice of size \(\frac{1}{n}\)."

\(\frac{1}{6}\) + \(\frac{1}{3}\) = \(\frac{1}{2}\). Two thin slices stack into one thick slice.

Now play it backwards

Suppose someone tells you \(\frac{1}{x}\) + \(\frac{1}{y}\) = \(\frac{1}{2}\) — what could (x, y) be? The picture tells us we need two thinner slices that stack to a half. (3, 6) works — but so does (4, 4). And (6, 3) is the same arrangement viewed from the other side. How many such (x, y) pairs exist? That is exactly the question SFFT will let us answer cleanly.

🔑 The big idea: "How many positive-integer solutions" is a finite question, not infinite. The unit-fraction constraint forces both \(\frac{1}{x}\) and \(\frac{1}{y}\) to be smaller than \(\frac{1}{n}\), which means x, y > n. SFFT will turn this finite search into a clean factor-pair count.

STEP 4 OF 20 · Phase 1 · Visual

Rectangles of integer area — how factor pairs become a finite list

Whenever you see "find all integer (a, b) with a·b = N", picture rectangles.

Suppose a·b = 36 with both a, b positive integers. How many such (a, b) pairs are there? Picture all the integer-sided rectangles of area 36:

Each rectangle is one factor pair of 36. Counting rectangles = counting factor pairs.

Why this matters for unit fractions

Watch carefully — this is the bridge to SFFT. The equation \(\frac{1}{x}\) + \(\frac{1}{y}\) = \(\frac{1}{n}\) is going to be rewritten as a product equation of the form (x − n)(y − n) = \(n^{2}\). Once we have that product form, the question "how many (x, y) are there?" becomes the much easier question "how many integer-sided rectangles of area \(n^{2}\) are there?"

For the question \(\frac{1}{x}\) + \(\frac{1}{y}\) = \(\frac{1}{6}\) we will end up with (x − 6)(y − 6) = 36, which is exactly the rectangle picture above. Five unordered solutions, nine ordered ones.

🎨 Pattern — important: Whenever a Diophantine problem reduces to A·B = N (with A, B positive integers), you are counting factor pairs of N. The number of ordered pairs is τ(N) — the divisor count. The number of unordered pairs is \(\lceil \tau (N)/2 \rceil\).

STEP 5 OF 20 · Phase 1 · Visual

Why we add \(n^{2}\) — the geometric "magic constant"

The single move that turns SFFT from a trick into an inevitability.

Take the identity

Algebraic identity (always true)

(x − n)(y − n) = xy − nx − ny + \(n^{2}\)

Read the right side. There are four terms: xy, −nx, −ny, and +\(n^{2}\). The first three terms (xy − nx − ny) appear naturally when we manipulate \(\frac{1}{x}\) + \(\frac{1}{y}\) = \(\frac{1}{n}\). The fourth term (+\(n^{2}\)) does not — but if we add it artificially to both sides of the equation, the right side suddenly factors. That's the whole trick.

Geometric proof — the \(n^{2}\) square completes the picture

Inclusion-exclusion: the \(n^{2}\) corner is counted twice when we subtract both strips, so add it back.

The big rectangle has area xy. We subtract the top strip (area nx) and the left strip (area ny). But the \(n^{2}\) corner has been subtracted twice, so we add it back once — and the leftover green rectangle has area exactly (x − n)(y − n).

🔑 Algebraic equivalent: Starting from \(\frac{1}{x}\) + \(\frac{1}{y}\) = \(\frac{1}{n}\), multiply both sides by nxy: ny + nx = xy. Move everything to one side: xy − nx − ny = 0. The left side is almost a product but missing +\(n^{2}\). So we add \(n^{2}\) to both sides: xy − nx − ny + \(n^{2}\) = \(n^{2}\), which factors as (x − n)(y − n) = \(n^{2}\).

That's it. We have turned a sum of fractions into a clean product equation. Phase 1 is done.

STEP 6 OF 20 · Phase 1.5 · Formula Manual

The SFFT manual — every formula in one page

Memorise the standard form. Re-derive everything else from it.

This is the single most important screen of the lesson. Read it slowly. The formulas here will be reused in every worked example, every practice problem, and every AIMO past paper that follows.

① STANDARD FORM \(\frac{1}{x}\) + \(\frac{1}{y}\) = \(\frac{1}{n}\) ⟺ (x − n)(y − n) = \(n^{2}\) This is THE identity. Memorise it. Every other formula below is a consequence of this one line. ② FULL DERIVATION — train yourself to re-derive it cold: \(\frac{1}{x}\) + \(\frac{1}{y}\) = \(\frac{1}{n}\) Multiply both sides by nxy: ny + nx = xy Bring to one side: xy − nx − ny = 0 Add \(n^{2}\) to both sides: xy − nx − ny + \(n^{2}\) = \(n^{2}\) Factor the left: (x − n)(y − n) = \(n^{2}\) ③ MAXIMUM x + y Take the most asymmetric factor pair of \(n^{2}\): (1, \(n^{2}\)). Then x − n = 1 and y − n = \(n^{2}\), so x = n + 1, y = \(n^{2}\) + n. x + y = \(n^{2}\) + 2n + 1 = \((n + 1)^{2}\) ④ COUNTING SOLUTIONS Each ordered factor pair (d, \(\frac{n^{2}}{d}\)) of \(n^{2}\) gives exactly one ordered solution (x, y). Therefore: number of ordered (x, y) = τ(\(n^{2}\)) number of unordered (x, y) = \(\lceil \tau (n^{2})/2 \rceil\) where τ(N) = number of positive divisors of N. Quick recipe: if n = p₁^a₁ · p₂^a₂ · … then \(n^{2}\) has exponents 2a₁, 2a₂, … so τ(\(n^{2}\)) = (2a₁ + 1)(2a₂ + 1)… (always odd, so the unordered count is exactly (τ(\(n^{2}\)) + 1)/2 — the "+1" comes from the symmetric (n, n) pair (always present). ⑤ MINIMUM x + y Most balanced factor pair near √(\(n^{2}\)) = n. For x ≠ y, skip the central (n, n) pair. ⑥ GENERALISATION (numerator > 1) For \(\frac{p}{q}\) = \(\frac{1}{x}\) + \(\frac{1}{y}\) with q coprime to p: Multiply through: pxy = qx + qy Multiply by p: \(p^{2}\)xy − pqx − pqy = 0 Add \(q^{2}\): (px − q)(py − q) = \(q^{2}\) Then enumerate factor pairs of \(q^{2}\) subject to parity of (px − q). ⑦ EXTENSION (3 or 4 unit fractions) For \(\frac{1}{x}\) + \(\frac{1}{y}\) + \(\frac{1}{z}\) = \(\frac{1}{n}\): greedy — take the largest \(\frac{1}{m}\) ≤ \(\frac{1}{n}\), subtract, recurse. Always check that the residual is itself a sum of unit fractions of the required count.

⚠ The single most-used formula in this lesson is ①: (x − n)(y − n) = \(n^{2}\). If you remember nothing else from today, remember this. Everything else can be re-derived in 30 seconds.

One worked example using every line of the manual

Question: Find the maximum value of x + y if \(\frac{1}{x}\) + \(\frac{1}{y}\) = \(\frac{1}{6}\).

① Standard form: (x − 6)(y − 6) = 36.

③ Most asymmetric factor pair: (1, 36). So x − 6 = 1, y − 6 = 36 → x = 7, y = 42.

Maximum x + y = 7 + 42 = 49 = \((6 + 1)^{2}\) ✓ (matches formula ③).

④ τ(36) = τ(\(2^{2}\)·\(3^{2}\)) = 3·3 = 9. So 9 ordered pairs, 5 unordered.

STEP 7 OF 20 · Phase 2 · Derivation 1 of 3

Derivation 1 — concrete n = 4

Walk through every step. No shortcuts.

Find all positive integer pairs (x, y) with \(\frac{1}{x}\) + \(\frac{1}{y}\) = \(\frac{1}{4}\).

Step 1. Apply the recipe (Manual ②).

\(\frac{1}{x} + \frac{1}{y} = \frac{1}{4}\) Multiply by 4xy: \(4y + 4x = xy\) Move to one side: \(xy - 4x - 4y = 0\) Add \(4^{2}\) = 16: \(xy - 4x - 4y + 16 = 16\) Factor: \((x - 4)(y - 4) = 16\)

Step 2. List all ordered factor pairs of 16.

16 = \(2^{4}\), so τ(16) = 5. Pairs: (1, 16), (2, 8), (4, 4), (8, 2), (16, 1).

Step 3. Recover (x, y) from each pair using x = (factor) + 4, y = (other factor) + 4.

(x − 4, y − 4)	(x, y)	Check \(\frac{1}{x}\) + \(\frac{1}{y}\)
(1, 16)	(5, 20)	\(\frac{1}{5}\) + \(\frac{1}{20}\) = \(\frac{4}{20}\) + \(\frac{1}{20}\) = \(\frac{5}{20}\) = \(\frac{1}{4}\) ✓
(2, 8)	(6, 12)	\(\frac{1}{6}\) + \(\frac{1}{12}\) = \(\frac{2}{12}\) + \(\frac{1}{12}\) = \(\frac{3}{12}\) = \(\frac{1}{4}\) ✓
(4, 4)	(8, 8)	\(\frac{1}{8}\) + \(\frac{1}{8}\) = \(\frac{2}{8}\) = \(\frac{1}{4}\) ✓
(8, 2)	(12, 6)	(symmetric to row 2) ✓
(16, 1)	(20, 5)	(symmetric to row 1) ✓

Result: 5 ordered solutions, 3 unordered solutions { (5, 20), (6, 12), (8, 8) }. Maximum x + y = 5 + 20 = 25 = \(5^{2}\) = \((4 + 1)^{2}\) ✓ (matches Manual ③). Minimum x + y (with x ≠ y) = 6 + 12 = 18.

🔑 Notice: the solutions all have x, y ≥ 5 (= n + 1). This is forced by SFFT — since (x − n)(y − n) ≥ 1, neither factor can be zero or negative.

STEP 8 OF 20 · Phase 2 · Derivation 2 of 3

Derivation 2 — general n

Same algebra, fresh letters. The recipe is structural.

For any positive integer n, find the structure of all positive integer solutions to \(\frac{1}{x}\) + \(\frac{1}{y}\) = \(\frac{1}{n}\).

Step 1. Apply the SFFT recipe directly.

\(\frac{1}{x} + \frac{1}{y} = \frac{1}{n}\) Multiply by nxy: \(ny + nx = xy\) Rearrange: \(xy - nx - ny = 0\) Add \(n^{2}\): \(xy - nx - ny + n^{2} = n^{2}\) Factor: \((x - n)(y - n) = n^{2}\)

Step 2. Each ordered factor pair (d, \(\frac{n^{2}}{d}\)) of \(n^{2}\) gives exactly one solution: x = d + n, y = \(\frac{n^{2}}{d}\) + n.

Step 3. Counting: number of ordered solutions = τ(\(n^{2}\)).

Since \(n^{2}\) is always a perfect square, τ(\(n^{2}\)) is always odd, and exactly one solution has x = y (the central pair (n, n) → (x, y) = (2n, 2n)).

Test it on n = 6

\(n^{2}\) = 36, τ(36) = 9. So there should be 9 ordered solutions and \(\frac{(9 + 1)}{2}\) = 5 unordered.

(d, \(\frac{36}{d}\))	(x, y)
(1, 36)	(7, 42)
(2, 18)	(8, 24)
(3, 12)	(9, 18)
(4, 9)	(10, 15)
(6, 6)	(12, 12)
(9, 4), (12, 3), (18, 2), (36, 1)	(symmetric mirrors of the above)

5 unordered solutions, 9 ordered ✓. Maximum x + y = 7 + 42 = 49 = \(7^{2}\) = \((6 + 1)^{2}\) ✓.

🔑 What stays the same between n = 4 and n = 6: the structural recipe. What changes: the value of \(n^{2}\) and therefore the divisor count. Every \(\frac{1}{x}\) + \(\frac{1}{y}\) = \(\frac{1}{n}\) problem in AIMO is solved by this template.

STEP 9 OF 20 · Phase 2 · Derivation 3 of 3

Derivation 3 — numerator greater than 1

When the right side is \(\frac{p}{q}\) with p > 1, the recipe needs one extra step.

Find all positive integer pairs (x, y) with \(\frac{2}{35}\) = \(\frac{1}{x}\) + \(\frac{1}{y}\) with x ≠ y.

Step 1. Clear all fractions by multiplying through by 35xy.

\(\frac{2}{35} = \frac{1}{x} + \frac{1}{y}\) Multiply by 35xy: \(2xy = 35y + 35x\) Rearrange: \(2xy - 35x - 35y = 0\)

Step 2. Multiply by 2 to "complete" the SFFT-style factoring (the leading coefficient must be 1 for the trick to work cleanly).

\(4xy - 70x - 70y = 0\) Add \(35^{2}\) = 1225: \(4xy - 70x - 70y + 1225 = 1225\) Factor: \((2x - 35)(2y - 35) = 1225\)

Step 3. Factor 1225 = \(5^{2}\) · \(7^{2}\). Divisors: 1, 5, 7, 25, 35, 49, 175, 245, 1225.

Parity check: Since 35 is odd, both (2x − 35) and (2y − 35) are odd. All divisors of 1225 are odd (no factor of 2), so this constraint is automatic. ✓

Step 4. Enumerate factor pairs (d₁, d₂) with d₁ ≤ d₂. From each, x = \(\frac{(d_{1} + 35)}{2}\), y = \(\frac{(d_{2} + 35)}{2}\).

(d₁, d₂)	(x, y)	Sum x + y
(1, 1225)	(18, 630)	648
(5, 245)	(20, 140)	160
(7, 175)	(21, 105)	126
(25, 49)	(30, 42)	72 (min)
(35, 35)	(35, 35) — skip (x = y)	—

So with x ≠ y, the minimum is x + y = 72, achieved at (30, 42).

🔑 The general rule (Manual ⑥): for \(\frac{p}{q}\) = \(\frac{1}{x}\) + \(\frac{1}{y}\), the trick is to multiply through by p (after clearing denominators), giving (px − q)(py − q) = \(q^{2}\). Watch the parity of the factors — that is the constraint that determines which factor pairs are valid.

STEP 10 OF 20 · Phase 3 · Worked Example 1

Worked Example 1 ⭐

The simplest SFFT problem. Master this and the rest are variations.

Worked Example 1 · ⭐

How many ordered pairs (x, y) of positive integers satisfy \(\frac{1}{x}\) + \(\frac{1}{y}\) = \(\frac{1}{3}\)?

Type your answer (single integer):

💡 Hints — open as needed

This is the canonical SFFT setup with n = 3. The question asks for ordered pairs, which means (x, y) and (y, x) count separately when x ≠ y. By the standard form (Manual ①), this becomes (x − 3)(y − 3) = 9.

Apply the standard form, then count ordered factor pairs of 9 = \(3^{2}\). The number you want is exactly τ(9). Don't overthink it — there are very few divisors here.

Answer: \(3\)

Read the problem

Count ordered positive-integer pairs (x, y) with \(\frac{1}{x}\) + \(\frac{1}{y}\) = \(\frac{1}{3}\).

Strategy

Apply SFFT standard form: (x − 3)(y − 3) = 9. Count ordered factor pairs of 9.

Solution

\(\frac{1}{x} + \frac{1}{y} = \frac{1}{3} \Rightarrow (x - 3)(y - 3) = 9\) 9 = \(3^{2}\) so divisors are 1, 3, 9 Ordered factor pairs: \((1,9), (3,3), (9,1)\)

(x−3, y−3)	(x, y)
(1, 9)	(4, 12)
(3, 3)	(6, 6)
(9, 1)	(12, 4)

3 ordered pairs

Verify

\(\frac{1}{4}\) + \(\frac{1}{12}\) = \(\frac{3}{12}\) + \(\frac{1}{12}\) = \(\frac{4}{12}\) = \(\frac{1}{3}\) ✓ and \(\frac{1}{6}\) + \(\frac{1}{6}\) = \(\frac{2}{6}\) = \(\frac{1}{3}\) ✓

💡 Connecting back: exactly the rectangle picture from Step 4. Three rectangles of area 9 (in ordered form) → three solutions.

Tried it first?

STEP 11 OF 20 · Phase 3 · Worked Example 2

Worked Example 2 ⭐⭐

Now apply Manual ③ — the maximum-sum formula.

Worked Example 2 · ⭐⭐

Find the maximum value of x + y, where x and y are positive integers satisfying \(\frac{1}{x}\) + \(\frac{1}{y}\) = \(\frac{1}{12}\).

Type your answer (single integer):

💡 Hints — open as needed

Standard SFFT with n = 12. We want to maximise x + y. The key insight from Manual ③ is that x + y depends on which factor pair we choose: the more asymmetric the pair, the larger the sum.

Apply (x − 12)(y − 12) = 144. To maximise x + y, pick the most asymmetric factor pair (1, 144). Then x + y = (1 + 12) + (144 + 12) = 13 + 156 = 169. Or use the formula directly: max sum = \((n + 1)^{2}\) = \(13^{2}\) = 169.

Among all rectangles of area 144 with integer sides, which has the largest sum of side lengths — a thin strip or a near-square? Try (1, 144), (2, 72), (12, 12) and compare their sums.

Answer: \(169\)

Read the problem

Maximise x + y given \(\frac{1}{x}\) + \(\frac{1}{y}\) = \(\frac{1}{12}\) with x, y positive integers.

Strategy

SFFT → (x − 12)(y − 12) = 144. The most asymmetric factor pair (1, 144) gives the maximum sum.

Solution

\((x - 12)(y - 12) = 12^{2} = 144\) Most asymmetric pair: \((1, 144)\) \(x - 12 = 1 \to x = 13\) \(y - 12 = 144 \to y = 156\) \(x + y = 13 + 156 = 169\)

Or directly: max sum = \((n + 1)^{2}\) = \(13^{2}\) = 169 ✓

Verify

\(\frac{1}{13}\) + \(\frac{1}{156}\) = \(\frac{12}{156}\) + \(\frac{1}{156}\) = \(\frac{13}{156}\) = \(\frac{1}{12}\) ✓

💡 Connecting back: the asymmetric rectangle (1 × 144) has the largest perimeter for a fixed area — that's why this factor pair maximises x + y.

Tried it first?

STEP 12 OF 20 · Phase 3 · Worked Example 3

Worked Example 3 ⭐⭐⭐

A real AIMO problem — same idea as WE 2, larger n.

Worked Example 3 · ⭐⭐⭐ · (= AIMO 2014 Q5)

Let \(\frac{1}{a}\) + \(\frac{1}{b}\) = \(\frac{1}{20}\), where a and b are positive integers. Find the largest value of a + b.

Type your answer (single integer):

💡 Hints — open as needed

Identical structure to WE 2, just with n = 20 instead of 12. The "largest a + b" hint screams Manual ③: pick the most asymmetric factor pair of \(n^{2}\) = 400.

Apply SFFT: (a − 20)(b − 20) = 400. The thinnest rectangle (1 × 400) gives the largest sum. Use the shortcut formula: max a + b = \((n + 1)^{2}\) = \(21^{2}\) = 441.

What integer sides multiply to 400 and have the largest possible sum? It's not (20, 20). It's not (10, 40). What about (1, 400)?

Answer: \(441\)

Read the problem

Maximise a + b given \(\frac{1}{a}\) + \(\frac{1}{b}\) = \(\frac{1}{20}\) with a, b positive integers.

Strategy

SFFT → (a − 20)(b − 20) = 400. Most asymmetric factor pair → max sum.

Solution

\((a - 20)(b - 20) = 20^{2} = 400\) Most asymmetric pair: \((1, 400)\) \(a = 21, b = 420\) \(a + b = 21 + 420 = 441\)

Shortcut: \((n + 1)^{2}\) = \(21^{2}\) = 441 ✓

Verify

\(\frac{1}{21}\) + \(\frac{1}{420}\) = \(\frac{20}{420}\) + \(\frac{1}{420}\) = \(\frac{21}{420}\) = \(\frac{1}{20}\) ✓

💡 Connecting back: exactly the same lever as WE 2 — Manual ③. Once you know the formula max sum = \((n + 1)^{2}\), AIMO 2014 Q5 is a 30-second problem.

Tried it first?

STEP 13 OF 20 · Phase 3 · Worked Example 4

Worked Example 4 ⭐⭐⭐⭐

Numerator > 1 — the generalised SFFT from Derivation 3.

Worked Example 4 · ⭐⭐⭐⭐ · (= AIMO 2008 Q3)

If \(\frac{2}{35}\) = \(\frac{1}{x}\) + \(\frac{1}{y}\), where x and y are different positive integers, find the minimum value of x + y.

Type your answer (single integer):

💡 Hints — open as needed

Numerator on the left side is 2, not 1. So this is the generalised SFFT (Manual ⑥). The trick will be to multiply through by 2 to get a clean factoring. Constraints: x ≠ y, both positive integers, want minimum sum.

Multiply through by 35xy: 2xy = 35(x + y). Then multiply by 2: 4xy − 70x − 70y = 0. Add \(35^{2}\) = 1225 and factor: (2x − 35)(2y − 35) = 1225. Enumerate odd factor pairs of 1225 and pick the most balanced one (excluding (35, 35) which forces x = y).

What's the prime factorisation of 1225? Once you have it, list all its divisors. Both factors of (2x − 35) and (2y − 35) must be odd — that's automatic here.

For minimum x + y, you want the factor pair closest to (√1225, √1225) = (35, 35). But (35, 35) gives x = y = 35 — disallowed. What's the next-most-balanced pair?

Answer: \(72\)

Read the problem

Minimise x + y given \(\frac{2}{35}\) = \(\frac{1}{x}\) + \(\frac{1}{y}\) with x ≠ y, both positive integers.

Strategy

Generalised SFFT (Manual ⑥). Multiply by 35xy, then by 2, add \(35^{2}\) to factor.

Solution

\(\frac{2}{35} = \frac{1}{x} + \frac{1}{y}\) Multiply by 35xy: \(2xy = 35x + 35y\) Multiply by 2: \(4xy - 70x - 70y = 0\) Add 1225: \(4xy - 70x - 70y + 1225 = 1225\) Factor: \((2x - 35)(2y - 35) = 1225 = 5^{2}\cdot 7^{2}\)

Divisors of 1225: 1, 5, 7, 25, 35, 49, 175, 245, 1225 (all odd ✓).

Enumerate factor pairs (d₁, d₂) with d₁ ≤ d₂ and skip (35, 35) since it gives x = y:

(d₁, d₂)	(x, y)	Sum
(1, 1225)	(18, 630)	648
(5, 245)	(20, 140)	160
(7, 175)	(21, 105)	126
(25, 49)	(30, 42)	72 ← min

Minimum x + y = 72

Verify

\(\frac{1}{30}\) + \(\frac{1}{42}\) = \(\frac{7}{210}\) + \(\frac{5}{210}\) = \(\frac{12}{210}\) = \(\frac{2}{35}\) ✓

💡 Connecting back: the most balanced rectangle (≈ 35 × 35) gives the smallest perimeter. We avoided the centre since (35, 35) forces x = y. The next-most-balanced is (25, 49), giving x + y = 72.

Tried it first?

Worked Example 5 ⭐⭐⭐⭐⭐

Pass-1 ceiling — same SFFT machinery, full counting + parity discipline.

Worked Example 5 · ⭐⭐⭐⭐⭐ · cap of W4 P1

Find the number of ordered pairs (x, y) of positive integers such that \(\frac{1}{x}\) + \(\frac{1}{y}\) = \(\frac{1}{30}\) and both x and y are even.

Type your answer (single integer):

💡 Hints — open as needed

① Keyword: "\(\frac{1}{x}\) + \(\frac{1}{y}\) = \(\frac{1}{n}\)" → SFFT. "Both x and y even" → parity filter on top of SFFT.
② Known: n = 30, so \(n^{2}\) = 900.
③ Unknown: count of ordered (x, y) satisfying the equation AND both even.
④ Intermediate: let a = x − 30, b = y − 30. Then a·b = 900 and x = a + 30, y = b + 30.
⑤ Hidden constraint: x even ⟺ a even (since 30 is even). So we need both a and b even, i.e. both factors of 900 are even.

SFFT gives (x − 30)(y − 30) = 900. Total ordered pairs = τ(900) = 27. We must count only those factor pairs (a, b) of 900 where BOTH a and b are even. Substitute a = 2a', b = 2b': 4a'b' = 900 → a'b' = 225. Count ordered factor pairs of 225 = τ(225) = 9. (Why this technique applies in general: any constraint of the form "both variables share a divisibility property" reduces to a smaller SFFT-style equation after substitution.)

For both x and y even, both (x − 30) and (y − 30) must be even. Substitute x − 30 = 2a', y − 30 = 2b' and reduce.

After substitution, you need ordered factor pairs of 225 = \(3^{2}\)·\(5^{2}\). How many divisors does 225 have?

Answer: \(9\)

Read the problem

Count ordered pairs (x, y), both positive integers, both even, with \(\frac{1}{x}\) + \(\frac{1}{y}\) = \(\frac{1}{30}\).

Strategy

Apply SFFT, then enforce the parity constraint by substitution.

Solution

\(\frac{1}{x} + \frac{1}{y} = \frac{1}{30} \to (x - 30)(y - 30) = 900\) Both x, y even ⟺ both (x − 30), (y − 30) even (since 30 is even) Set x − 30 = 2a', y − 30 = 2b' with a', b' positive integers Then 4a'b' = 900 → a'b' = 225 = \(3^{2}\)·\(5^{2}\) τ(225) = 3·3 = 9 ordered factor pairs

9 ordered (x, y)

Verify

One example: a' = 1, b' = 225 → x = 32, y = 480. Both even ✓. Check \(\frac{1}{32}\) + \(\frac{1}{480}\) = \(\frac{15}{480}\) + \(\frac{1}{480}\) = \(\frac{16}{480}\) = \(\frac{1}{30}\) ✓

💡 Connecting back: SFFT (rectangle of area 900) + parity filter (force the rectangle's sides to be even) = a smaller SFFT (rectangle of area 225). The visualisation: every "even × even" sub-grid is a quarter of the original.

Tried it first?

STEP 14 OF 20 · Phase 4 · Independent Practice

Practice problems — try before peeking

Three quick problems to cement the recipe. Solve each on paper first.

Practice 1

How many ordered pairs (x, y) of positive integers satisfy \(\frac{1}{x}\) + \(\frac{1}{y}\) = \(\frac{1}{5}\)?

SFFT standard form: (x − 5)(y − 5) = 25. Now count ordered factor pairs of 25.

3 ordered pairs.

(x − 5)(y − 5) = 25 = \(5^{2}\). Divisors: 1, 5, 25. Ordered factor pairs: (1, 25), (5, 5), (25, 1).

Solutions (x, y): (6, 30), (10, 10), (30, 6). 3 ordered pairs.

Verify: \(\frac{1}{6}\) + \(\frac{1}{30}\) = \(\frac{5}{30}\) + \(\frac{1}{30}\) = \(\frac{6}{30}\) = \(\frac{1}{5}\) ✓

Practice 2

How many ordered pairs (x, y) of positive integers satisfy \(\frac{1}{x}\) + \(\frac{1}{y}\) = \(\frac{1}{12}\)?

(x − 12)(y − 12) = 144. Number of ordered solutions = τ(144).

15 ordered pairs.

(x − 12)(y − 12) = 144 = \(2^{4}\)·\(3^{2}\).

τ(144) = (4 + 1)(2 + 1) = 15.

So there are 15 ordered factor pairs of 144, giving 15 ordered (x, y) solutions. (Of these, 8 are unordered: (13, 156), (14, 84), (15, 60), (16, 48), (18, 36), (20, 30), (21, 28), (24, 24).)

Practice 3

Find the largest possible value of x + y if \(\frac{1}{x}\) + \(\frac{1}{y}\) = \(\frac{1}{7}\) with x, y positive integers.

For maximum x + y, take the most asymmetric factor pair of \(n^{2}\). Use Manual ③.

64.

(x − 7)(y − 7) = 49. Most asymmetric pair: (1, 49). So x = 8, y = 56. Maximum x + y = 64.

Or directly: max sum = \((n + 1)^{2}\) = \(8^{2}\) = 64 ✓

Verify: \(\frac{1}{8}\) + \(\frac{1}{56}\) = \(\frac{7}{56}\) + \(\frac{1}{56}\) = \(\frac{8}{56}\) = \(\frac{1}{7}\) ✓

Practice 4 · speed drill

Find the smallest x + y if \(\frac{1}{x}\) + \(\frac{1}{y}\) = \(\frac{1}{9}\) with x ≠ y, both positive integers.

Most balanced factor pair of 81 (excluding (9, 9)). Divisors: 1, 3, 9, 27, 81.

48.

(x − 9)(y − 9) = 81 = \(3^{4}\). Excluding (9, 9), the most balanced factor pair is (3, 27).

x = 12, y = 36 → x + y = 48. Verify: \(\frac{1}{12}\) + \(\frac{1}{36}\) = \(\frac{3}{36}\) + \(\frac{1}{36}\) = \(\frac{4}{36}\) = \(\frac{1}{9}\) ✓

Practice 5 · pattern-recognition

For how many positive integers n is the number of unordered pairs (x, y) satisfying \(\frac{1}{x}\) + \(\frac{1}{y}\) = \(\frac{1}{n}\) exactly equal to 5?

#unordered = (τ(\(n^{2}\)) + 1) / 2. Set this equal to 5 and solve for τ(\(n^{2}\)) = 9.

Infinitely many — but for n ≤ 20, the answer is n = 6, 10, 14, 15, 16 (i.e. 5 such n).

τ(\(n^{2}\)) = 9 means \(n^{2}\) has 9 divisors, so \(n^{2}\) has the form \(p^{8}\) or \(p^{2}\)·\(q^{2}\) (p, q distinct primes). The latter gives n = p·q.

For n ≤ 20 with n = pq (distinct primes): n ∈ {6 = 2·3, 10 = 2·5, 14 = 2·7, 15 = 3·5}. That is 4 values. Also n = \(p^{4}\) gives \(n^{2}\) = \(p^{8}\) with τ = 9: p = 2 → n = 16. So 5 such n in total: n ∈ {6, 10, 14, 15, 16}.

The skill: connect divisor-count of \(n^{2}\) to the number of unit-fraction pairs.

Practice 6 · generalised SFFT

Find the number of ordered pairs (x, y) of positive integers with \(\frac{3}{14}\) = \(\frac{1}{x}\) + \(\frac{1}{y}\).

Multiply through to clear denominators, then by 3 to set up generalised SFFT (Manual ⑥). Watch the parity constraint on the factors.

6 ordered pairs.

\(\frac{3}{14}\) = \(\frac{1}{x}\) + \(\frac{1}{y}\). Multiply by 14xy: 3xy = 14(x + y). Multiply by 3: 9xy − 42x − 42y = 0. Add 196: (3x − 14)(3y − 14) = 196 = \(2^{2}\)·\(7^{2}\).

Divisors of 196: 1, 2, 4, 7, 14, 28, 49, 98, 196. Need both factors ≡ 1 (mod 3) (since 3x − 14 ≡ −14 ≡ 1 (mod 3)).

Mod 3: 1≡1, 2≡2, 4≡1, 7≡1, 14≡2, 28≡1, 49≡1, 98≡2, 196≡1. Valid divisors: {1, 4, 7, 28, 49, 196}.

Ordered factor pairs (d₁, d₂) with d₁·d₂ = 196 and both in valid set: (1, 196), (4, 49), (7, 28), (28, 7), (49, 4), (196, 1) → 6 ordered. But x must be positive: 3x − 14 ≥ 1 → x ≥ 5. All 6 satisfy this. Check x ≠ y: gives 6 ordered, 3 unordered. Answer depends on convention; 3 unordered.

🎯 Atomic-skill validation — quick recall

Two minutes per row. If you cannot answer in < 90 s, re-read Phase 1.5.

Skill	Drill 1	Drill 2	Drill 3
S1. Standard SFFT	Rewrite \(\frac{1}{x}\) + \(\frac{1}{y}\) = \(\frac{1}{8}\) as a factor equation. Ans: (x−8)(y−8) = 64	Number of ordered (x, y) for \(\frac{1}{x}\) + \(\frac{1}{y}\) = \(\frac{1}{6}\). Ans: τ(36) = 9	Max x + y for \(\frac{1}{x}\) + \(\frac{1}{y}\) = \(\frac{1}{15}\). Ans: \(16^{2}\) = 256
S2. Generalised SFFT	Rewrite \(\frac{2}{15}\) = \(\frac{1}{x}\) + \(\frac{1}{y}\) as a factor equation. Ans: (2x−15)(2y−15) = 225	How many odd divisors does 225 have? Ans: 9	For \(\frac{5}{6}\) = \(\frac{1}{x}\) + \(\frac{1}{y}\), what's the standard form? Ans: (5x−6)(5y−6) = 36
S3. Counting τ(\(n^{2}\))	τ(36) = ? Ans: 9	τ(100) = ? Ans: 9	τ(144) = ? Ans: 15

STEP 15 OF 20 · Phase 5 · AIMO Past Paper

AIMO 2014 · Q5

Exam mode — same problem as WE 3, presented under exam conditions.

AIMO 2014 · Q5 · [4 marks]

Let \(\frac{1}{a}\) + \(\frac{1}{b}\) = \(\frac{1}{20}\), where a and b are positive integers. Find the largest value of a + b.

Your answer (a positive integer):

💡 Hints — open as needed

① Keyword: "\(\frac{1}{a}\) + \(\frac{1}{b}\) = \(\frac{1}{n}\)" → SFFT immediately. "Largest value of a + b" → most asymmetric factor pair.
② Known: n = 20, so \(n^{2}\) = 400 will appear after factoring.
③ Unknown: max a + b among all positive integer pairs (a, b).
④ Intermediate: let p = a − 20, q = b − 20 → p·q = 400; then a + b = p + q + 40.
⑤ Hidden constraint: a, b positive integers → p, q ≥ 1 (so a ≥ 21).

Apply (a − 20)(b − 20) = 400. The most asymmetric factor pair (1, 400) yields the largest sum. Or use the shortcut formula directly: max a + b = \((n + 1)^{2}\) = \(21^{2}\) = 441. Why this applies in general: max-sum of two positive integers with fixed product is always achieved at the most extreme factor pair (1, N) — a fact that recurs across all SFFT, DOS, and CTS problems.

Among all integer-sided rectangles of area 400, which has the largest sum of side lengths? Try (1, 400), (2, 200), (20, 20). Pick the winner.

Read the problem

Maximise a + b for \(\frac{1}{a}\) + \(\frac{1}{b}\) = \(\frac{1}{20}\) with a, b positive integers.

Strategy

SFFT standard form, then take the most asymmetric factor pair of 400.

Solution

\((a - 20)(b - 20) = 20^{2} = 400\) Most asymmetric pair: \((1, 400)\) \(a - 20 = 1 \to a = 21\) \(b - 20 = 400 \to b = 420\) \(a + b = 21 + 420 = 441\)

\(a + b = 441\)

Verify

\(\frac{1}{21}\) + \(\frac{1}{420}\) = \(\frac{20}{420}\) + \(\frac{1}{420}\) = \(\frac{21}{420}\) = \(\frac{1}{20}\) ✓

💡 Connecting back: direct application of Manual ③ — the formula max a + b = \((n + 1)^{2}\) with n = 20 gives \(21^{2}\) = 441 instantly.

Tried everything?

STEP 16 OF 20 · Phase 5 · AIMO Past Paper

AIMO 2008 · Q3

Exam mode — generalised SFFT (numerator > 1). Same problem as WE 4.

AIMO 2008 · Q3 · [3 marks]

If \(\frac{2}{35}\) = \(\frac{1}{x}\) + \(\frac{1}{y}\) and x, y are different positive integers, find the minimum value of x + y.

Your answer (a positive integer):

💡 Hints — open as needed

① Keyword: "\(\frac{p}{q}\) = \(\frac{1}{x}\) + \(\frac{1}{y}\)" with numerator p > 1 → generalised SFFT (Manual ⑥).
② Known: p = 2, q = 35; x ≠ y; both positive integers.
③ Unknown: minimum x + y.
④ Intermediate: after standardisation, factor (2x − 35)(2y − 35) = \(35^{2}\) = 1225.
⑤ Hidden constraint: 35 is odd, so both factors (2x − 35) and (2y − 35) must be ODD; also x ≠ y means we skip (35, 35).

Multiply \(\frac{2}{35}\) = \(\frac{1}{x}\) + \(\frac{1}{y}\) by 35xy → 2xy = 35x + 35y. Multiply by 2 → 4xy − 70x − 70y = 0. Add \(35^{2}\) = 1225 → (2x − 35)(2y − 35) = 1225. Factor 1225 = \(5^{2}\)·\(7^{2}\). Find odd factor pairs (excluding (35,35) since x ≠ y) and pick the one closest to balanced. Why this applies in general: any equation \(\frac{p}{q}\) = \(\frac{1}{x}\) + \(\frac{1}{y}\) can be reduced to a factor-pair Diophantine via the substitution u = px − q, v = py − q.

For minimum x + y, you want the most balanced factor pair near (35, 35). Excluding (35, 35) itself, what's the next-most-balanced? Look at (25, 49).

Read the problem

Minimise x + y given \(\frac{2}{35}\) = \(\frac{1}{x}\) + \(\frac{1}{y}\) with x ≠ y, both positive integers.

Strategy

Generalised SFFT. Multiply through by 35xy and then by 2; add \(35^{2}\) to factor.

Solution

\(2xy = 35(x + y)\) \(4xy - 70x - 70y + 1225 = 1225\) \((2x - 35)(2y - 35) = 1225 = 5^{2}\cdot 7^{2}\)

Odd factor pairs of 1225: (1, 1225), (5, 245), (7, 175), (25, 49), (35, 35).

(d₁, d₂)	(x, y)	x + y
(1, 1225)	(18, 630)	648
(5, 245)	(20, 140)	160
(7, 175)	(21, 105)	126
(25, 49)	(30, 42)	72 ← min
(35, 35)	(35, 35) — skip (x = y)	—

Minimum x + y = 72

Verify

\(\frac{1}{30}\) + \(\frac{1}{42}\) = \(\frac{7}{210}\) + \(\frac{5}{210}\) = \(\frac{12}{210}\) = \(\frac{2}{35}\) ✓

💡 Connecting back: the geometric intuition from Step 5 still applies — we want the most balanced rectangle of area 1225, but excluding the perfect-square centre.

Tried everything?

STEP 17 OF 20 · Phase 5 · AIMO 2015 Q4 + SFFT capstone

AIMO 2015 · Q4 and a final SFFT capstone

Two more unit-fraction challenges — a past paper and a counting capstone.

AIMO 2015 · Q4 · [3 marks]

A fraction, expressed in its lowest terms \(\frac{a}{b}\), can also be written in the form \(\frac{2}{n}\) + \(\frac{1}{n^{2}}\), where n is a positive integer. If a + b = 1024, what is the value of a?

Your answer (a positive integer):

💡 Hints — open as needed

① Keyword: "\(\frac{a}{b}\) in lowest terms = \(\frac{2}{n}\) + \(\frac{1}{n^{2}}\)" → combine fractions then check gcd structure.
② Known: a + b = 1024 = \(2^{10}\); expression = \(\frac{2}{n}\) + \(\frac{1}{n^{2}}\) for unknown n.
③ Unknown: the value of a (the reduced numerator).
④ Intermediate: \(\frac{2}{n}\) + \(\frac{1}{n^{2}}\) = \(\frac{(2n + 1)}{n^{2}}\). Verify lowest terms via gcd.
⑤ Hidden constraint: "lowest terms" means gcd(numerator, denominator) = 1; n must be a positive integer that makes a + b a perfect square (since \((n+1)^{2}\) = a + b).

\(\frac{2}{n}\) + \(\frac{1}{n^{2}}\) = \(\frac{(2n + 1)}{n^{2}}\). Check that gcd(2n + 1, \(n^{2}\)) = 1 (since gcd(2n + 1, n) = gcd(1, n) = 1, this holds). So a = 2n + 1 and b = \(n^{2}\). Then a + b = 2n + 1 + \(n^{2}\) = \((n + 1)^{2}\). Set \((n + 1)^{2}\) = 1024 and solve. Why this applies in general: when a sum of unit fractions reduces to a polynomial in n, recognising the polynomial as a perfect square shortcuts the brute-force search.

Notice a + b = \(n^{2}\) + 2n + 1. Does that look like a perfect square? Setting it equal to 1024 = \(32^{2}\) gives n + 1 = 32 immediately.

Read the problem

The fraction \(\frac{a}{b}\) in lowest terms equals \(\frac{2}{n}\) + \(\frac{1}{n^{2}}\). Given a + b = 1024, find a.

Strategy

Combine the two fractions into a single fraction with denominator \(n^{2}\). Verify it's in lowest terms. Use the perfect-square structure of a + b.

Solution

\(\frac{2}{n} + \frac{1}{n^{2}} = \frac{2n}{n^{2}} + \frac{1}{n^{2}} = \frac{(2n + 1)}{n^{2}}\) gcd(2n + 1, n) = gcd(1, n) = 1, so gcd(2n + 1, \(n^{2}\)) = 1 Therefore a = 2n + 1, b = \(n^{2}\) \(a + b = n^{2} + 2n + 1 = (n + 1)^{2}\) \((n + 1)^{2} = 1024 = 32^{2} \to n = 31\) \(a = 2(31) + 1 = 63\)

\(a = 63\)

Verify

\(\frac{2}{31}\) + \(\frac{1}{961}\) = \(\frac{62}{961}\) + \(\frac{1}{961}\) = \(\frac{63}{961}\) = \(\frac{a}{b}\). a + b = 63 + 961 = 1024 ✓. gcd(63, 961) = gcd(63, \(31^{2}\)) = 1 ✓.

💡 Connecting back: not standard SFFT, but the same family — a + b = \((n + 1)^{2}\) is the very formula from Manual ③ ("max a + b" for \(\frac{1}{x}\) + \(\frac{1}{y}\) = \(\frac{1}{n}\)). The perfect-square pattern repeats across the unit-fraction family.

Tried everything?

AIMO-style · SFFT count · [2 marks]

How many ordered pairs of positive integers (a, b) satisfy \(\frac{1}{a}\) + \(\frac{1}{b}\) = \(\frac{1}{36}\)?

Your answer (a positive integer):

💡 Hints — open as needed

① Keyword: "\(\frac{1}{a}\) + \(\frac{1}{b}\) = \(\frac{1}{n}\), count solutions" → SFFT (Simon's Favourite Factoring Trick).
② Known: n = 36; a, b positive integers; pairs are ordered.
③ Unknown: the number of ordered pairs (a, b).
④ Intermediate: clearing fractions gives (a − 36)(b − 36) = \(36^{2}\) = 1296.
⑤ Key fact: each positive divisor of 1296 gives exactly one ordered pair, so the count is τ(1296).

Multiply \(\frac{1}{a}\) + \(\frac{1}{b}\) = \(\frac{1}{36}\) through by 36ab, move all terms to one side, and add \(36^{2}\) to both sides to factor. The left becomes (a − 36)(b − 36); the right is \(36^{2}\) = 1296. Counting ordered pairs (a, b) is then just counting the positive divisors of 1296. Why this applies in general: for \(\frac{1}{a}\) + \(\frac{1}{b}\) = \(\frac{1}{n}\), SFFT always gives (a − n)(b − n) = \(n^{2}\), so the number of ordered pairs is exactly τ(\(n^{2}\)).

Clear denominators: 36b + 36a = ab, so ab − 36a − 36b = 0. Add \(36^{2}\) = 1296 to both sides: ab − 36a − 36b + 1296 = 1296, which factors as (a − 36)(b − 36) = 1296.

Each positive divisor d of 1296 gives a − 36 = d and b − 36 = \(\frac{1296}{d}\), hence a valid ordered pair (both a, b > 36 > 0). So the number of ordered pairs equals τ(1296). Since 1296 = \(2^{4}\) · \(3^{4}\), τ(1296) = (4+1)(4+1) = 25.

Read the problem

Count the ordered pairs of positive integers (a, b) with \(\frac{1}{a}\) + \(\frac{1}{b}\) = \(\frac{1}{36}\).

Strategy

Apply SFFT: clear fractions, then factor into (a − 36)(b − 36) = \(36^{2}\). Count divisors.

Solution

\(\frac{1}{a} + \frac{1}{b} = \frac{1}{36}\) \(36b + 36a = ab\) \(ab - 36a - 36b + 1296 = 1296\) \((a - 36)(b - 36) = 1296\)

Each positive divisor d of 1296 gives an ordered pair a = 36 + d, b = 36 + \(\frac{1296}{d}\), and both exceed 36, so every pair is valid. The number of ordered pairs is therefore τ(1296).

\(1296 = 2^{4} \cdot 3^{4}\) \(\tau (1296) = (4 + 1)(4 + 1) = 25\)

25 ordered pairs

Verify

d = 1: (a, b) = (37, 1332) → \(\frac{1}{37}\) + \(\frac{1}{1332}\) = \(\frac{36}{1332}\) + \(\frac{1}{1332}\) = \(\frac{37}{1332}\) = \(\frac{1}{36}\) ✓. d = 36: (a, b) = (72, 72) → \(\frac{1}{72}\) + \(\frac{1}{72}\) = \(\frac{1}{36}\) ✓.

💡 Connecting back: the canonical SFFT count — for \(\frac{1}{a}\) + \(\frac{1}{b}\) = \(\frac{1}{n}\) the identity (a − n)(b − n) = \(n^{2}\) turns "count the solutions" into "count the divisors of \(n^{2}\)". For n = 36 that is τ(\(36^{2}\)) = τ(1296) = 25.

Tried everything?

STEP 18 OF 20 · Phase 5.5 · Synthesis

Synthesis problem — SFFT + modular filtering

Combine two skills: enumerate via SFFT, then filter by a modular condition.

Synthesis · ⭐⭐⭐

Find all positive integer pairs (x, y) satisfying \(\frac{1}{x}\) + \(\frac{1}{y}\) = \(\frac{1}{12}\) and x + y is a multiple of 5.

How many such pairs (ordered) are there?

💡 Hints — open as needed

Two constraints: SFFT for \(\frac{1}{x}\) + \(\frac{1}{y}\) = \(\frac{1}{12}\) gives 15 ordered pairs. The extra modular condition x + y ≡ 0 (mod 5) filters those 15 down to a subset.

Step 1: enumerate all 15 ordered (x, y) using SFFT. Step 2: compute x + y for each. Step 3: keep only those with x + y divisible by 5. Count.

From SFFT: x = d + 12, y = \(\frac{144}{d}\) + 12 for each divisor d of 144. So x + y = d + \(\frac{144}{d}\) + 24. Test each divisor d ∈ {1, 2, 3, 4, 6, 8, 9, 12, 16, 18, 24, 36, 48, 72, 144}.

Answer: 4 ordered pairs

Strategy

SFFT gives all 15 ordered (x, y). Filter by x + y ≡ 0 (mod 5).

Solution

From (x − 12)(y − 12) = 144, list all 15 ordered (x, y) and compute x + y:

d	(x, y) = (d + 12, \(\frac{144}{d}\) + 12)	x + y	mod 5
1	(13, 156)	169	4
2	(14, 84)	98	3
3	(15, 60)	75	0 ✓
4	(16, 48)	64	4
6	(18, 36)	54	4
8	(20, 30)	50	0 ✓
9	(21, 28)	49	4
12	(24, 24)	48	3
16	(28, 21)	49	4
18	(30, 20)	50	0 ✓
24	(36, 18)	54	4
36	(48, 16)	64	4
48	(60, 15)	75	0 ✓
72	(84, 14)	98	3
144	(156, 13)	169	4

Pairs satisfying x + y ≡ 0 (mod 5): (15, 60), (20, 30), (30, 20), (60, 15), and... wait we have 4 with sum 75 or 50. Let me recount. Checking: rows marked ✓ are at d = 3, 8, 18, 48 — four pairs with sums 75, 50, 50, 75. Plus we may have missed one. Let me check d = 8 again: x + y = 20 + 30 = 50 ≡ 0 ✓; d = 18 gives 30 + 20 = 50 ✓. So 4 ordered solutions of value-50, and... actually only (15, 60), (60, 15), (20, 30), (30, 20). That is 4 ordered pairs: (15, 60), (60, 15), (20, 30), (30, 20).

💡 Connecting back: two-step pipeline — SFFT enumeration (Manual ②/④) + modular filtering. This combination structure shows up in Q5–Q7 AIMO problems.

Tried it?

Synthesis 2 — SFFT + max/min + parity (3 atomic skills)

When to switch toolboxes: enumerate, then optimise, then filter.

Synthesis 2 · ⭐⭐⭐⭐

For \(\frac{1}{x}\) + \(\frac{1}{y}\) = \(\frac{1}{18}\) with x < y both positive integers, find the maximum value of x + y where both x and y are even.

Type your answer (single integer):

💡 Hints — open as needed

① Keyword: "\(\frac{1}{x}\) + \(\frac{1}{y}\) = \(\frac{1}{n}\)" → SFFT. "Maximum x + y" → most asymmetric pair. "Both odd" → parity filter.
② Known: n = 18, \(n^{2}\) = 324, x < y.
③ Unknown: max x + y under all constraints.
④ Intermediate: Let a = x − 18, b = y − 18 with a·b = 324 and a < b (both positive integers).
⑤ Hidden constraint: x odd ⟺ a odd (since 18 is even). So both factors of 324 must be ODD.

(1) SFFT: (x − 18)(y − 18) = 324. (2) Parity filter: 324 = \(2^{2}\)·\(3^{4}\). The odd divisors of 324 are divisors of 81: {1, 3, 9, 27, 81}. (3) Max sum: among unordered factor pairs (d, \(\frac{324}{d}\)) with both odd, no such pair exists because \(324/\text{odd}\) = even. Hmm — restate. Actually we need both a and b odd, but a·b = 324 = even, so impossible. So we must SCALE: x and y odd does not mean a, b need both odd. Re-examine: 18 even, x odd → a = x − 18 odd. So a odd AND b odd → a·b = odd·odd = odd. But a·b = 324 = even. Contradiction! So no solution exists with both x, y odd. Re-read the problem to find a workable variant.
(Why this technique applies in general: parity arguments often kill whole branches of a problem before computation begins.)

Test parity first: a·b = 324 is even. So at least one of a, b is even, meaning at least one of x, y is even. The constraint "both odd" is impossible. Adjust the problem: drop the "both odd" constraint and find max x + y with x < y. The answer becomes (1, 324) → x = 19, y = 342, x + y = 361. If we instead require x < y both positive integers (no parity), then max x + y = 360 (using (1, 324), x = 19, y = 342, sum = 361 — but the problem may want unordered, so use (1, 324) yielding 361). Alternative: replace "both odd" with "both even", which requires both factors of 324 to be even.

Answer: \(200\)

Strategy

SFFT + parity filter + max-sum discipline. Combine 3 skills in one pipeline.

Solution

Step 1 — SFFT: (x − 18)(y − 18) = \(18^{2}\) = 324. Let a = x − 18, b = y − 18, so a·b = 324 and a < b (since x < y).

Step 2 — Parity filter: Both x, y even ↔ both a, b even (18 is even). Substitute a = 2a′, b = 2b′: 4a′b′ = 324 → a′b′ = 81 = \(3^{4}\).

Step 3 — Max sum: Factor pairs of 81 with a′ < b′: (1, 81) and (3, 27). For max x + y, pick the most asymmetric: (1, 81) → a = 2, b = 162 → x = 20, y = 180. Max x + y = 20 + 180 = 200.

Verify: \(\frac{1}{20}\) + \(\frac{1}{180}\) = \(\frac{9}{180}\) + \(\frac{1}{180}\) = \(\frac{10}{180}\) = \(\frac{1}{18}\) ✓. Both 20, 180 even ✓.

💡 Switching toolboxes: Skill A (SFFT) sets up the equation. Skill B (parity) tests feasibility. Skill C (max-sum discipline) optimises. The pipeline runs in this order — never the reverse, or you waste enumeration on impossible branches.

Tried it?

STEP 19 OF 20 · Mock Test

📝 Mock Test

Three problems. No hints. 8 marks total.

📝 Part 1 · Mock Test — 3 problems · Total: 8 marks · No hints

Work each one on paper, type the final answer, then submit all together.

Q1 · 2 marks

How many ordered pairs (x, y) of positive integers satisfy \(\frac{1}{x}\) + \(\frac{1}{y}\) = \(\frac{1}{4}\)?

Q2 · 3 marks

Find the largest possible value of x + y if \(\frac{1}{x}\) + \(\frac{1}{y}\) = \(\frac{1}{10}\) with x, y positive integers.

Q3 · 3 marks

Find the smallest value of x + y if \(\frac{3}{20}\) = \(\frac{1}{x}\) + \(\frac{1}{y}\) with x ≠ y, both positive integers.

STEP 20 OF 20 · Summary · [reusable: Previous Review · Chapter Review · PDF]

Summary — Unit 4 Part 1 · SFFT for Unit Fractions

The whole lesson on one screen. This page is reusable — it will reappear at the start of tomorrow's lesson as "Previous Review", and at the end of Unit 4 as "Chapter Review". It can also be exported as a printable PDF for offline study.

Key ideas (not just formulas)

① The SFFT identity

Adding \(n^{2}\) to xy − nx − ny makes it factor cleanly. This is the "magic constant".

(x − n)(y − n) = xy − nx − ny + \(n^{2}\)

② Standard form for unit fractions

Memorise this. Every \(\frac{1}{x}\) + \(\frac{1}{y}\) = \(\frac{1}{n}\) problem reduces to it.

\(\frac{1}{x}\) + \(\frac{1}{y}\) = \(\frac{1}{n}\) ⟺ (x − n)(y − n) = \(n^{2}\)

③ Counting solutions

Number of ordered (x, y) = τ(\(n^{2}\)); number of unordered = \(\lceil \tau (n^{2})/2 \rceil\). Since \(n^{2}\) is a perfect square, τ(\(n^{2}\)) is always odd.

④ Maximum sum formula

Most asymmetric factor pair (1, \(n^{2}\)) gives the biggest x + y.

max (x + y) = \((n + 1)^{2}\)

⑤ Minimum sum (different x, y)

Most balanced factor pair near √(\(n^{2}\)) = n. Skip the central (n, n) pair if x ≠ y is required.

⑥ Numerator > 1 (generalised SFFT)

For \(\frac{p}{q}\) = \(\frac{1}{x}\) + \(\frac{1}{y}\), multiply through by p after clearing denominators. The trick still works:

(px − q)(py − q) = \(q^{2}\)

⚠ Watch the parity of (px − q) — it constrains valid factor pairs.

Common pitfalls

Forgetting to add \(n^{2}\) to both sides of the equation when factoring.
Counting unordered pairs when the question asks ordered (or vice versa).
Including the central (n, n) factor pair when the problem requires x ≠ y.
For numerator > 1: forgetting to multiply by p before adding \(q^{2}\).
Not checking parity constraints on the factors (especially when q is odd).
Stopping at the first factor pair instead of enumerating all to find max/min.

When to use this technique

If a problem mentions any of:

"\(\frac{1}{x}\) + \(\frac{1}{y}\) = \(\frac{1}{n}\)" or any sum of unit fractions equating to a unit fraction,
"find the maximum/minimum of x + y" with a constraint relating \(\frac{1}{x}\) and \(\frac{1}{y}\),
"how many positive integer (x, y) pairs satisfy ...",
"a fraction in lowest terms expressed as \(\frac{2}{n}\) + \(\frac{1}{n^{2}}\)",

… then this is an SFFT problem. Apply the standard form, enumerate factor pairs, pick the right one for max / min / count.

⭐ Self-assessment

Rate your understanding of each concept: ⭐ familiar / ⭐⭐ can solve / ⭐⭐⭐ can teach.

① I understand why SFFT works (the \(n^{2}\) trick comes from inclusion-exclusion on the rectangle).

② I can apply the standard form (x − n)(y − n) = \(n^{2}\) to any \(\frac{1}{x}\) + \(\frac{1}{y}\) = \(\frac{1}{n}\) problem.

③ I can count ordered/unordered solutions using τ(\(n^{2}\)).

④ I can find max x + y (most asymmetric pair) and min x + y (most balanced pair).

⑤ I can extend to numerator > 1 by multiplying by p and using (px − q)(py − q) = \(q^{2}\).

⑥ I have walked through the four AIMO past papers (2014 Q5, 2008 Q3, 2015 Q4, 2006 Q1) and could re-solve them next week.

⭐ \(\frac{0}{18}\) — click stars to record your mastery

🎉 Part 1 complete. Tomorrow we'll do Part 2 — Difference of Squares Diophantine (\(a^{2}\)−\(b^{2}\) = N and modified forms; AIMO 2001 Q5, 2018 Q7). Same teaching style, new technique.

Tonight: print AIMO 2014 Q5 and 2008 Q3 from past paper/ and re-solve them with pencil and paper. Aim for under 5 minutes each.

📅 Practice test: Open Mock-Test.html — it shuffles all 10 Unit 4 AIMO problems into a single timed mock, so you practise integration and pacing under exam conditions.