You may also like

problem icon

Tweedle Dum and Tweedle Dee

Two brothers were left some money, amounting to an exact number of pounds, to divide between them. DEE undertook the division. "But your heap is larger than mine!" cried DUM...

problem icon

Lower Bound

What would you get if you continued this sequence of fraction sums? 1/2 + 2/1 = 2/3 + 3/2 = 3/4 + 4/3 =

problem icon

Sum Equals Product

The sum of the numbers 4 and 1 [1/3] is the same as the product of 4 and 1 [1/3]; that is to say 4 + 1 [1/3] = 4 1 [1/3]. What other numbers have the sum equal to the product and can this be so for any whole numbers?

Unit Fractions

Stage: 3 Challenge Level: Challenge Level:3 Challenge Level:3 Challenge Level:3

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