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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... ### 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 ï¿½ 1 [1/3]. What other numbers have the sum equal to the product and can this be so for any whole numbers? ### Special Sums and Products

Find some examples of pairs of numbers such that their sum is a factor of their product. eg. 4 + 12 = 16 and 4 × 12 = 48 and 16 is a factor of 48.

# Unit Fractions

##### Age 11 to 14Challenge Level

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.

#### 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