Unit fractions
Consider the equation 1/a + 1/b + 1/c = 1 where a, b and c are
natural numbers and 0 < a < b < c. Prove that there is
only one set of values which satisfy this equation.
Age
11 to 14
Challenge level
Problem
Consider the equation ${1\over a} +{1\over b}+{1\over c} = 1$ where $a$, $b$ and $c$ are natural numbers and $0 < a < b < c$. Prove that $a< 3$ and also that $b< 4$ and hence that there is only one set of values which satisfy this equation.
Find the six sets of values which satisfy the equation ${1\over a} +{1\over b}+{1\over c} +{1\over d}= 1$ where $a$, $b$, $c$ and $d$ are natural numbers and $0 < a < b < c < d$.
Student Solutions
Here is Andrei Lazanu's solution. Andrei is 12 years old and a
pupil at School Number 205, Bucharest, Romania. You don't need much
mathematical knowledge to solve this problem but you do need a lot
of mathematical thinking. Well done Andrei.
"I started from the relation $0< a< b< c$, so $a$,
$b$ and $c$ are positive. This means that: $${1 \over a} > {1
\over b} > {1\over c}$$Consequently, the original equality: $${1
\over a} + {1 \over b} + {1\over c} = 1$$transforms, keeping into
account (1) (${1\over a}> {1\over b}$ and ${1\over a}>
{1\over c}$) into: $${3\over a} > 1$$ i.e. $a< 3$.
Because $a$ is integer, and because it cannot be 1 (if it
would, then: ${1\over b} + {1\over c}=0$) it means that the only
one possibility is $a = 2$. Then: $${1\over 2}+{1\over b}+{1\over
c} = 1 \mbox{ or } {1\over b}+{1\over c} = {1\over 2}$$Because from
(1): ${1\over b} > {1\over c}$, (3) transforms into: ${2\over
b}> {1\over 2}$ or $b < 4$. Because $b > a$ and $a = 2$,
there is only one possibility for $b$, $b = 3$. Substituting into
the original relation, I obtained $c = 6$.
With four numbers, I followed the same procedure. From:
${1\over a} + {1\over b} +{1\over c} + {1\over d } = 1$
I obtained: ${4\over a}> 1$, so $a < 4$. Again $a$
cannot be 1, so it must be either 2 or 3. I analyse the
possibilities one by one.
a = 2
Using again the relation of order between $b$, $c$ and $d$ I found $b < 6$. Because $b > a$, I have to analyse $b = 3$, $b = 4$ and $b = 5$.
b = 3
From the relation between $c$ and $d$, and substituting the values for $a$ and $b$ I obtained $c < 12$, it means it could be 4, 5, 6, 7, 8, 9, 10 or 11.
c = 4
${1\over 2}+{1\over 3}+{1\over 4}+{1\over d} = 1$ has no solution for $d$ integer.
c = 5
${1\over 2}+{1\over 3}+{1\over 5}+{1\over d} = 1$ has no solution for $d$ integer.
c = 6
${1\over 2}+{1\over 3}+{1\over 6}+{1\over d} = 1$ has no solution for $d$ integer.
c = 7
${1\over 2}+{1\over 3}+{1\over 7}+{1\over d} = 1$ gives $d=42$.
c = 8
${1\over 2}+{1\over 3}+{1\over 8}+{1\over d} = 1$ gives $d=24$.
c = 9
${1\over 2}+{1\over 3}+{1\over 9}+{1\over d} = 1$ gives $d=18$.
c = 10
${1\over 2}+{1\over 3}+{1\over 10}+{1\over d} = 1$ gives $d=15$.
c = 11
${1\over 2}+{1\over 3}+{1\over 11}+{1\over d} = 1$ has no solution for $d$ integer.
b = 4
Repeating the same procedure, I obtained: $c < 8$, and because $c > b$, $c$ could be 5, 6 and 7.
c = 5
${1\over 2}+{1\over 4}+{1\over 5}+{1\over d} = 1$ gives $d=20$.
c = 6
${1\over 2}+{1\over 4}+{1\over 6}+{1\over d} = 1$ gives $d=12$.
c = 7
${1\over 2}+{1\over 4}+{1\over 7}+{1\over d} = 1$ hasn't a solution for $d$ integer.
b = 5
I obtained: $3c < 20$, and consequently only $c =6$ is possible. This hasn't a solution for $d$.
a = 3
For $b$, I obtained: $2b < 9$, and because $b > a$, it is possible only $b = 4$. Then, I obtained $5c < 24$, i. e. without solution because $c > b$.
Finally, the good solutions are:
| a | 2 | 2 | 2 | 2 | 2 | 2 |
| b | 3 | 3 | 3 | 3 | 4 | 4 |
| c | 7 | 8 | 9 | 10 | 5 | 6 |
| d | 42 | 24 | 18 | 15 | 20 | 12 |