Question 2

We need to find four different unit fractions that add up to 1.

Step 1 — Finding the first fraction

Let us call the four fractions $1/a$, $1/b$, $1/c$, and $1/d$. We want $1/a + 1/b + 1/c + 1/d = 1$. The denominators $a, b, c, d$ must all be different. To make the sum 1, one of the fractions must be large. The largest possible unit fraction is $1/2$. If the smallest denominator is 3, then the sum of the four smallest different unit fractions starting with $1/3$ would be $1/3 + 1/4 + 1/5 + 1/6$. Let us add these fractions. The common bottom number for $3, 4, 5, 6$ is 60.

$\frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \frac{1}{6}$

$= \frac{20}{60} + \frac{15}{60} + \frac{12}{60} + \frac{10}{60}$

$= \frac{20 + 15 + 12 + 10}{60}$

$= \frac{57}{60}$

This sum is less than 1. So, one of the fractions must be $1/2$. Let $a = 2$.

Step 2 — Finding the remaining three fractions

Now we have $1/2 + 1/b + 1/c + 1/d = 1$. This means $1/b + 1/c + 1/d = 1 - 1/2$.

$1 - \frac{1}{2}$

$= \frac{2}{2} - \frac{1}{2}$

$\boxed{\frac{1}{2}}$

So we need three different unit fractions $1/b, 1/c, 1/d$ that add up to $1/2$. The denominators $b, c, d$ must be different from each other and from 2. Let us choose the next smallest denominator for $b$. If $b=3$, then $1/3 + 1/c + 1/d = 1/2$. This means $1/c + 1/d = 1/2 - 1/3$.

$\frac{1}{2} - \frac{1}{3}$

$= \frac{3}{6} - \frac{2}{6}$

$\boxed{\frac{1}{6}}$

So we need two different unit fractions $1/c$ and $1/d$ that add up to $1/6$. The denominators $c, d$ must be different from 2 and 3. Also, $1/c$ must be smaller than $1/6$, so $c$ must be greater than 6. Let us try $c = 7$. Then $1/d = 1/6 - 1/7$.

$\frac{1}{6} - \frac{1}{7}$

$= \frac{7}{42} - \frac{6}{42}$

$\boxed{\frac{1}{42}}$

So $d = 42$. This gives our first solution: $1/2 + 1/3 + 1/7 + 1/42 = 1$. The denominators are $2, 3, 7, 42$. They are all different.

Step 3 — Finding more solutions

We continue to find $1/c + 1/d = 1/6$. We need $c > 6$ and $c \ne d$. If $c = 8$: Then $1/d = 1/6 - 1/8$.

$\frac{1}{6} - \frac{1}{8}$

$= \frac{4}{24} - \frac{3}{24}$

$\boxed{\frac{1}{24}}$

So $d = 24$. This gives our second solution: $1/2 + 1/3 + 1/8 + 1/24 = 1$. The denominators are $2, 3, 8, 24$. They are all different.

If $c = 9$: Then $1/d = 1/6 - 1/9$.

$\frac{1}{6} - \frac{1}{9}$

$= \frac{3}{18} - \frac{2}{18}$

$\boxed{\frac{1}{18}}$

So $d = 18$. This gives our third solution: $1/2 + 1/3 + 1/9 + 1/18 = 1$. The denominators are $2, 3, 9, 18$. They are all different.

If $c = 10$: Then $1/d = 1/6 - 1/10$.

$\frac{1}{6} - \frac{1}{10}$

$= \frac{5}{30} - \frac{3}{30}$

$= \frac{2}{30}$

$\boxed{\frac{1}{15}}$

So $d = 15$. This gives our fourth solution: $1/2 + 1/3 + 1/10 + 1/15 = 1$. The denominators are $2, 3, 10, 15$. They are all different.

If $c = 11$: $1/6 - 1/11 = 5/66$, which is not a unit fraction. If $c = 12$: $1/6 - 1/12 = 1/12$. Here $c=d=12$, but $c$ and $d$ must be different. So we have found all solutions starting with $1/2 + 1/3$.

Step 4 — Exploring other options for the second fraction

We still need $1/b + 1/c + 1/d = 1/2$. We used $b=3$. Let us try $b=4$. Remember $b$ must be different from 2. So $1/4 + 1/c + 1/d = 1/2$. This means $1/c + 1/d = 1/2 - 1/4$.

$\frac{1}{2} - \frac{1}{4}$

$= \frac{2}{4} - \frac{1}{4}$

$\boxed{\frac{1}{4}}$

So we need two different unit fractions $1/c$ and $1/d$ that add up to $1/4$. The denominators $c, d$ must be different from 2 and 4. Also, $1/c$ must be smaller than $1/4$, so $c$ must be greater than 4. Let us try $c = 5$. Then $1/d = 1/4 - 1/5$.

$\frac{1}{4} - \frac{1}{5}$

$= \frac{5}{20} - \frac{4}{20}$

$\boxed{\frac{1}{20}}$

So $d = 20$. This gives our fifth solution: $1/2 + 1/4 + 1/5 + 1/20 = 1$. The denominators are $2, 4, 5, 20$. They are all different.

If $c = 6$: Then $1/d = 1/4 - 1/6$.

$\frac{1}{4} - \frac{1}{6}$

$= \frac{3}{12} - \frac{2}{12}$

$\boxed{\frac{1}{12}}$

So $d = 12$. This gives our sixth solution: $1/2 + 1/4 + 1/6 + 1/12 = 1$. The denominators are $2, 4, 6, 12$. They are all different.

If $c = 7$: $1/4 - 1/7 = 3/28$, which is not a unit fraction. If $c = 8$: $1/4 - 1/8 = 1/8$. Here $c=d=8$, but $c$ and $d$ must be different. So we have found all solutions starting with $1/2 + 1/4$.

Step 5 — Checking other options for the second fraction (b)

We still need $1/b + 1/c + 1/d = 1/2$. We used $b=3$ and $b=4$. Let us try $b=5$. Remember $b$ must be different from 2. So $1/5 + 1/c + 1/d = 1/2$. This means $1/c + 1/d = 1/2 - 1/5$.

$\frac{1}{2} - \frac{1}{5}$

$= \frac{5}{10} - \frac{2}{10}$

$\boxed{\frac{3}{10}}$

So we need two different unit fractions $1/c$ and $1/d$ that add up to $3/10$. The denominators $c, d$ must be different from 2 and 5. Also, $1/c$ must be smaller than $3/10$, so $c$ must be greater than $10/3$, which is about 3.33. So $c$ can be 4 or 6. (Cannot be 5 as it is already used). If $c = 4$: Then $1/d = 3/10 - 1/4$.

$\frac{3}{10} - \frac{1}{4}$

$= \frac{6}{20} - \frac{5}{20}$

$\boxed{\frac{1}{20}}$

So $d = 20$. This gives $1/2 + 1/5 + 1/4 + 1/20 = 1$. This is the same set of fractions as $1/2 + 1/4 + 1/5 + 1/20 = 1$. We already counted this.

If $c = 6$: Then $1/d = 3/10 - 1/6$.

$\frac{3}{10} - \frac{1}{6}$

$= \frac{9}{30} - \frac{5}{30}$

$= \frac{4}{30}$

$= \frac{2}{15}$

This is not a unit fraction. So $c=6$ does not work. We have found all six solutions.

Step 6 — Visualising one solution

Let us visualise the solution $1/2 + 1/4 + 1/5 + 1/20 = 1$. We can divide a circle into 20 equal parts. $1/2$ is $10/20$ of the circle. $1/4$ is $5/20$ of the circle. $1/5$ is $4/20$ of the circle. $1/20$ is $1/20$ of the circle. We can color these parts differently.

Answer

(i) One solution is $1/2 + 1/3 + 1/7 + 1/42 = 1$. (ii) The six solutions are: $1/2 + 1/3 + 1/7 + 1/42 = 1$ $1/2 + 1/3 + 1/8 + 1/24 = 1$ $1/2 + 1/3 + 1/9 + 1/18 = 1$ $1/2 + 1/3 + 1/10 + 1/15 = 1$ $1/2 + 1/4 + 1/5 + 1/20 = 1$ $1/2 + 1/4 + 1/6 + 1/12 = 1$