Ages: 11β14 Β· Duration: 90 minutes Β· Topics: Fractions, Number Theory, Factoring, Simon's Favorite Factoring Trick
"You have one pizza to share equally among 3 friends. Easy: each gets $\frac{1}{3}$. But what if your knife can only cut unit fractions β slices like $\frac{1}{2}$, $\frac{1}{4}$, $\frac{1}{6}$? Can you still give everyone a fair share?"
The ancient Egyptians faced exactly this problem. Their number system only had notation for fractions with 1 on top β what we call unit fractions: $\frac{1}{2}$, $\frac{1}{3}$, $\frac{1}{7}$, $\frac{1}{42}$, etc. To write any fraction, they had to break it into a sum of distinct unit fractions.
For example, instead of $\frac{2}{3}$, an Egyptian scribe would write:
$$\frac{2}{3} = \frac{1}{2} + \frac{1}{6}$$| Task | Hint | Answer |
|---|---|---|
| Write $\frac{2}{5}$ as a sum of two unit fractions | Try $\frac{1}{5} + \frac{1}{?}$ | $\frac{1}{3} + \frac{1}{15}$ |
| Write $\frac{2}{7}$ as a sum of two unit fractions | $\frac{1}{a} + \frac{1}{b}$ where $a \neq b$ | $\frac{1}{4} + \frac{1}{28}$ |
| Is $\frac{1}{6} + \frac{1}{6}$ a valid Egyptian fraction? | Must the parts be different? | No! The Egyptians required distinct denominators |
Before tackling the competition problem, let's warm up with a simpler equation.
Question: Find all pairs of distinct positive integers $(a, b)$ such that $$\frac{2}{15} = \frac{1}{a} + \frac{1}{b}$$
Step 1: Clear the fractions.
Start from $\frac{2}{15} = \frac{1}{a} + \frac{1}{b}$. Multiply both sides by $15ab$:
$$2ab = 15(a + b) = 15a + 15b$$Rearrange:
$$2ab - 15a - 15b = 0$$Step 2: Factor the left side. This is the tricky part! We want something of the form $(β¦)(β¦)$.
Multiply the whole equation by 2:
$$4ab - 30a - 30b = 0$$Now add $15^2 = 225$ to both sides:
$$4ab - 30a - 30b + 225 = 225$$The left side factors beautifully:
$$(2a - 15)(2b - 15) = 225$$Step 3: List factor pairs.
We need $(2a - 15)(2b - 15) = 225 = 3^2 \times 5^2$.
Since $a, b$ are positive integers and $2a - 15, 2b - 15$ must be odd (because 15 is odd), we list all positive odd factor pairs of 225:
| $2a - 15$ | $2b - 15$ | $a$ | $b$ | Distinct? | $a + b$ |
|---|---|---|---|---|---|
| 1 | 225 | 8 | 120 | β | 128 |
| 3 | 75 | 9 | 45 | β | 54 |
| 5 | 45 | 10 | 30 | β | 40 |
| 9 | 25 | 12 | 20 | β | 32 |
| 15 | 15 | 15 | 15 | β | β |
Step 4: Verify one solution. Take $(a, b) = (9, 45)$:
$$\frac{1}{9} + \frac{1}{45} = \frac{5}{45} + \frac{1}{45} = \frac{6}{45} = \frac{2}{15} \quad \checkmark$$FSJM Competition β Challenge 15: Egyptian Fractions
Matilda found this equality in an old notebook of her grandfather: $$\frac{2}{85} = \frac{1}{\boxed{?}} + \frac{1}{\boxed{?}}$$ Two different denominators are hidden by ink stains.
What is the sum of these two denominators? Find all the possibilities.
"We just cracked $\frac{2}{15}$. The method was: clear fractions, factor, list factor pairs. Can you apply the same idea to $\frac{2}{85}$? Turn to your neighbour and set it up."
Step 1: Clear fractions. From $\frac{2}{85} = \frac{1}{a} + \frac{1}{b}$, multiply by $85ab$:
$$2ab = 85a + 85b$$ $$2ab - 85a - 85b = 0$$Step 2: Apply SFFT. Multiply by 2 and add $85^2$:
$$4ab - 170a - 170b + 7225 = 7225$$ $$(2a - 85)(2b - 85) = 7225$$Step 3: Factorize 7225.
$$7225 = 85^2 = (5 \times 17)^2 = 5^2 \times 17^2$$"How many positive divisors does $5^2 \times 17^2$ have?"
The number of divisors is $(2+1)(2+1) = 9$. The divisors are:
$$1, \; 5, \; 17, \; 25, \; 85, \; 289, \; 425, \; 1{,}445, \; 7{,}225$$All are odd (no factor of 2), which is exactly what we need.
Step 4: List all factor pairs with $d_1 \leq d_2$ and $d_1 \cdot d_2 = 7225$:
| $d_1 = 2a{-}85$ | $d_2 = 2b{-}85$ | $a$ | $b$ | Distinct? | $a + b$ |
|---|---|---|---|---|---|
| 1 | 7225 | 43 | 3655 | β | 3698 |
| 5 | 1445 | 45 | 765 | β | 810 |
| 17 | 425 | 51 | 255 | β | 306 |
| 25 | 289 | 55 | 187 | β | 242 |
| 85 | 85 | 85 | 85 | β (same!) | β |
Five factor pairs, but one gives $a = b = 85$, which is forbidden. That leaves four valid decompositions.
A student might think: "I'll just try $a = 43, 44, 45, \ldots$ and see which give integer $b$."
From $\frac{1}{b} = \frac{2}{85} - \frac{1}{a}$, we get $b = \frac{85a}{2a - 85}$. You'd have to test every $a > 42$ (since $2a > 85$) and check whether $b$ is a positive integer β and you'd stopβ¦ when? At $a = 3655$?
This brute-force approach works, but it gives no guarantee you've found them all, and no insight into why there are exactly four. The factoring method is both complete and illuminating.
Intuition: We're "completing the rectangle" β just like completing the square, but for products of two different variables.
How we used it: Our equation $2ab - 85a - 85b = 0$ has $A = 2$, $B = -85$, $C = -85$, $D = 0$:
$$(2a - 85)(2b - 85) = 2 \cdot 0 + (-85)(-85) = 7225$$Why it matters: The trick turns a two-variable equation into a single factoring problem. Once you factor the right-hand side, you can enumerate all solutions β guaranteed.
Start from $Axy + Bx + Cy = D$.
Multiply both sides by $A$:
$$A^2xy + ABx + ACy = AD$$Add $BC$ to both sides:
$$A^2xy + ABx + ACy + BC = AD + BC$$Factor the left side by grouping:
$$Ax(Ay + B) + C(Ay + B) = AD + BC$$ $$(Ax + C)(Ay + B) = AD + BC \qquad \blacksquare$$For the equation $\frac{2}{n} = \frac{1}{a} + \frac{1}{b}$, the SFFT gives:
$$(2a - n)(2b - n) = n^2$$So the number of decompositions (with $a \neq b$) is:
$$\frac{\tau(n^2) - 1}{2}$$where $\tau(n^2)$ is the number of odd positive divisors of $n^2$ (subtracting 1 for the $a = b$ case, dividing by 2 because each pair is counted twice).
For $n = 85 = 5 \times 17$, we get $n^2 = 5^2 \times 17^2$, so $\tau(n^2) = 3 \times 3 = 9$ (all divisors are odd). The number of valid pairs is $\frac{9 - 1}{2} = 4$. β
| Problem | Answer |
|---|---|
| How many ways can $\frac{2}{21}$ be written as a sum of two distinct unit fractions? Hint: $21^2 = 441 = 3^2 \times 7^2$, so $\tau = 9$. | $\frac{9 - 1}{2} = 4$ ways |
| Find the pair with the smallest sum for $\frac{2}{21}$. | $(2a - 21)(2b - 21) = 441$; use $(9, 49)$: $a = 15, b = 35$, sum $= 50$ |
| For which values of $n$ does $\frac{2}{n}$ have a unique decomposition into two distinct unit fractions? | When $n^2$ has exactly 3 odd divisors, i.e., $n$ is an odd prime |
From $\frac{2}{85} = \frac{1}{a} + \frac{1}{b}$ with $a \neq b$ and $a, b \in \mathbb{Z}^+$:
$$2ab = 85(a + b) \implies (2a - 85)(2b - 85) = 85^2 = 7225$$The positive divisor pairs of $7225 = 5^2 \times 17^2$ with $d_1 < d_2$:
| $d_1$ | $d_2$ | $a = \frac{85 + d_1}{2}$ | $b = \frac{85 + d_2}{2}$ | $a + b$ |
|---|---|---|---|---|
| 1 | 7225 | 43 | 3655 | 3698 |
| 5 | 1445 | 45 | 765 | 810 |
| 17 | 425 | 51 | 255 | 306 |
| 25 | 289 | 55 | 187 | 242 |
| Pair $(a, b)$ | $\frac{1}{a} + \frac{1}{b}$ | $= \frac{2}{85}$? |
|---|---|---|
| $(43, 3655)$ | $\frac{3655 + 43}{43 \times 3655} = \frac{3698}{157{,}165}$ | $\frac{2}{85} = \frac{3698}{157{,}165}$ β |
| $(45, 765)$ | $\frac{765 + 45}{45 \times 765} = \frac{810}{34{,}425}$ | $\frac{2}{85} = \frac{810}{34{,}425}$ β |
| $(51, 255)$ | $\frac{255 + 51}{51 \times 255} = \frac{306}{13{,}005}$ | $\frac{2}{85} = \frac{306}{13{,}005}$ β |
| $(55, 187)$ | $\frac{187 + 55}{55 \times 187} = \frac{242}{10{,}285}$ | $\frac{2}{85} = \frac{242}{10{,}285}$ β |
Instead of the SFFT, we can solve directly for $b$ in terms of $a$.
From $\frac{2}{85} = \frac{1}{a} + \frac{1}{b}$:
$$\frac{1}{b} = \frac{2}{85} - \frac{1}{a} = \frac{2a - 85}{85a}$$ $$b = \frac{85a}{2a - 85}$$For $b$ to be a positive integer, we need:
Let $d = 2a - 85$, so $a = \frac{85 + d}{2}$. Then:
$$b = \frac{85(85 + d)}{2d} = \frac{7225 + 85d}{2d} = \frac{7225}{2d} + \frac{85}{2}$$For $b$ to be an integer, we need $d$ to be an odd positive divisor of $7225$. Since $d$ is automatically odd (as $85$ is odd and $2a$ is even), we simply need $d \mid 7225$.
The odd positive divisors of $7225$ are: $1, 5, 17, 25, 85, 289, 425, 1445, 7225$.
Excluding $d = 85$ (which gives $a = b = 85$), and noting that $d > 85$ gives $a > b$ (just swaps the pair), we get the same four solutions.
We can derive tight bounds for $a$ (assuming $a < b$) and then search.
From $\frac{2}{85} = \frac{1}{a} + \frac{1}{b}$ with $a < b$:
Range: $43 \leq a \leq 84$.
For each $a$ in $\{43, 44, \ldots, 84\}$, compute $b = \frac{85a}{2a - 85}$ and check if it's a positive integer:
| $a$ | $2a - 85$ | $\frac{85a}{2a-85}$ | Integer? |
|---|---|---|---|
| 43 | 1 | 3655 | β |
| 44 | 3 | $1246.\overline{6}$ | β |
| 45 | 5 | 765 | β |
| 46 | 7 | $558.57\ldots$ | β |
| β¦ | β¦ | β¦ | β¦ |
| 51 | 17 | 255 | β |
| β¦ | β¦ | β¦ | β¦ |
| 55 | 25 | 187 | β |
| β¦ | β¦ | β¦ | β¦ |
Only 4 values of $a$ produce integer $b$: exactly $a \in \{43, 45, 51, 55\}$.
The ErdΕsβStraus conjecture (1948, still open!) states that for every integer $n \geq 2$:
$$\frac{4}{n} = \frac{1}{a} + \frac{1}{b} + \frac{1}{c}$$has a solution in positive integers. Our problem is a cousin: decomposing $\frac{2}{n}$ into two unit fractions (a simpler case that's fully solved).
General theory. The equation $\frac{2}{n} = \frac{1}{a} + \frac{1}{b}$ with $a \neq b$ and $a < b$ has solutions in bijection with factorizations $n^2 = d_1 \cdot d_2$ with $d_1 < d_2$, $d_1$ and $d_2$ having the same parity as $n$, via:
$$a = \frac{n + d_1}{2}, \qquad b = \frac{n + d_2}{2}$$When $n$ is odd (as $85$ is), every divisor is automatically odd, so we need all factorizations of $n^2$.
Counting formula. The number of valid pairs is:
$$N = \frac{\tau(n^2) - 1}{2}$$For $n = pq$ with $p, q$ distinct odd primes: $\tau(n^2) = \tau(p^2 q^2) = 3 \times 3 = 9$, giving $N = 4$.
Classification by structure. The four decompositions of $\frac{2}{85}$ can be classified by which factorization of $85 = 5 \times 17$ they exploit:
| Factor pair of $85^2$ | Decomposition | "Shape" |
|---|---|---|
| $1 \times 7225$ | $\frac{1}{43} + \frac{1}{3655}$ | Trivial split: $b \approx \frac{n^2}{2}$ |
| $5 \times 1445$ | $\frac{1}{45} + \frac{1}{765}$ | Factor-5 split |
| $17 \times 425$ | $\frac{1}{51} + \frac{1}{255}$ | Factor-17 split |
| $25 \times 289$ | $\frac{1}{55} + \frac{1}{187}$ | "Balanced" split |
The last pair $(55, 187)$ is the most balanced (closest to $a = b = 85$), while the first pair $(43, 3655)$ is the most extreme.
| Pair $(a, b)$ | Egyptian Fraction Decomposition | Sum $a + b$ |
|---|---|---|
| $(43, 3655)$ | $\frac{2}{85} = \frac{1}{43} + \frac{1}{3655}$ | $\boxed{3698}$ |
| $(45, 765)$ | $\frac{2}{85} = \frac{1}{45} + \frac{1}{765}$ | $\boxed{810}$ |
| $(51, 255)$ | $\frac{2}{85} = \frac{1}{51} + \frac{1}{255}$ | $\boxed{306}$ |
| $(55, 187)$ | $\frac{2}{85} = \frac{1}{55} + \frac{1}{187}$ | $\boxed{242}$ |
| Method | Key Idea | Prerequisites |
|---|---|---|
| Solution 1 β SFFT | Factor $(2a-85)(2b-85) = 7225$ | Factoring, divisibility |
| Solution 2 β Direct algebra | Solve $b = \frac{85a}{2a-85}$ and require $d \mid 7225$ | Fraction manipulation |
| Solution 3 β Bounded search | Derive $43 \leq a \leq 84$, test each | Inequalities, computation |
| Solution 4 β ErdΕsβStraus | Classify via $\tau(n^2)$ | Number theory, divisor function |
Problem 1: $\frac{2}{21} = \frac{1}{a} + \frac{1}{b}$ gives $(2a - 21)(2b - 21) = 441 = 3^2 \times 7^2$.
Factor pairs: $(1, 441), (3, 147), (7, 63), (9, 49)$, plus $(21, 21)$ excluded.
Solutions: $(11, 231)$, $(12, 84)$, $(14, 42)$, $(15, 35)$.
Sums: $242, 96, 56, 50$.
Problem 2: $\frac{1}{55} + \frac{1}{187} = \frac{187 + 55}{55 \times 187} = \frac{242}{10{,}285}$. And $\frac{2}{85} = \frac{242}{85 \times 121} = \frac{242}{10{,}285}$. β
Problem 3: For $n = p$ (odd prime), $\tau(p^2) = 3$, so there is $\frac{3-1}{2} = 1$ decomposition:
$(2a - p)(2b - p) = p^2$ with factor pair $(1, p^2)$, giving $a = \frac{p+1}{2}$ and $b = \frac{p(p+1)}{2}$.
Problem 4: The greedy algorithm picks $a = \lceil 85/2 \rceil = 43$ (the largest unit fraction $\leq \frac{2}{85}$), giving $b = 3655$. This produces the pair $(43, 3655)$ β the most "unbalanced" solution.
Problem 5: This is a significantly harder enumeration. A computer search finds the complete list (the number of solutions grows rapidly with the number of summands).
Problem 6: Each solution $(a, b)$ with $a < b$ corresponds uniquely to a factorization $d_1 \cdot d_2 = n^2$ with $0 < d_1 < d_2$ (both odd when $n$ is odd). There are $\frac{\tau(n^2) - 1}{2}$ such pairs (subtract 1 for $d_1 = d_2 = n$, divide by 2 for ordering). β