Unit Fractions

Age 11 to 14

Challenge Level Yellow star

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

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For younger learners

Unit Fractions

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:

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