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# Special Sums and Products

There were many good solutions to this one, and they are all special cases of the solution given by Syed Farhan Iskandar, age 12, of Foxford School and Community College. We are looking for a formula which gives pairs of numbers $a$ and $b$ such that $a + b$ is a factor of $ab$. Here is Farhan's method:

If the answer to the product divided by the sum is $s$ then

$\begin{eqnarray} \\ s(a + b) &=& ab \\ sa + sb &=& ab \\ sa &=& ab - sb = b(a - s) \\ b &=& \frac{sa}{a - s}\quad (1). \end{eqnarray}$

We can take any number $a$ and any whole number $s$ where $s \neq a$ and Farhan's formula gives a pair of numbers with the required property.

[Note that $a $ and $b$ do not have to be whole numbers and also that, if $s > a$, we introduce negative numbers, but we'll come back to that later.

Farhan gives the following examples:

$\begin{eqnarray} \\ a&=&2,\ s=1\ {\rm \ gives }\ b=2 \\ 2 + 2&=&4,\ 2\times 2\ =\ 4 \ {\rm and }\ 4/4 = 1 = s. \end{eqnarray}$

$\begin{eqnarray} \\ a&=&6,\ s=2 {\rm \ gives }\ b=3 \\ 6 + 3 &=&9,\ 6\times 3 = 18\ {\rm and }\ 18/9 = 2 = s. \end{eqnarray}$

$\begin{eqnarray} \\ a&=&12,\ s=3 {\rm \ gives }\ b=4 \\ 12 + 4 &=&16,\ 12\times 4 = 48\ {\rm and }\ 48/16 = 3 = s. \end{eqnarray}$

Davide Colli, age 13, Scuola Media Besozzi, who lives near Milan in Italy, showed that the number pair $a$ and $a(a - 1)$ is a solution for all values of $a$ (except $a = 1$). This is a special case of (1) with $s = a - 1$.

The Mount School York sent lots of examples taking the second number to be a multiple of the first. Charis Campbell, Christianne Eaves, Sheila Luk and Cheryl Wong proved that the pattern always works for two equal even numbers.

$$N + N = 2N, \ N\times N = N^2 \ {\rm and} \ N^2/2N = N/2$$

which is a whole number if and only if $N$ is even.

Sheila, Cheryl and Peach from year 10 gave the formula

$\begin{eqnarray} \\ a&=&A(k + 1),\ s = Ak, \ b=Ak(k + 1) \\ a + b &=& A(k + 1)^2, \ ab = A^2k(k + 1)^2 {\rm and }\ ab/(a + b) = Ak = s. \end{eqnarray}$

Again this is a special case of Farhan's result and it fits the pattern when $A$ and $k$ are whole numbers.

Farhan's formula given in (1) is not restricted to pairs of positive whole numbers as the following examples show:

$\begin{eqnarray} \\ a = 2, s=5 {\rm \ gives }\ b = (-10/3) \\ a + b = (-4/3),\ ab = (-20/3) \ {\rm and }\ ab/(a + b) = 5 = s. \end{eqnarray}$

$\begin{eqnarray} \\ a = 1/2, s=3 {\rm \ gives }\ b = (-3/5) \\ a + b = (-1/10),\ ab = (-3/10) \ {\rm and }\ ab/(a + b) = 3 = s. \end{eqnarray}$

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There were many good solutions to this one, and they are all special cases of the solution given by Syed Farhan Iskandar, age 12, of Foxford School and Community College. We are looking for a formula which gives pairs of numbers $a$ and $b$ such that $a + b$ is a factor of $ab$. Here is Farhan's method:

If the answer to the product divided by the sum is $s$ then

$\begin{eqnarray} \\ s(a + b) &=& ab \\ sa + sb &=& ab \\ sa &=& ab - sb = b(a - s) \\ b &=& \frac{sa}{a - s}\quad (1). \end{eqnarray}$

We can take any number $a$ and any whole number $s$ where $s \neq a$ and Farhan's formula gives a pair of numbers with the required property.

[Note that $a $ and $b$ do not have to be whole numbers and also that, if $s > a$, we introduce negative numbers, but we'll come back to that later.

Farhan gives the following examples:

$\begin{eqnarray} \\ a&=&2,\ s=1\ {\rm \ gives }\ b=2 \\ 2 + 2&=&4,\ 2\times 2\ =\ 4 \ {\rm and }\ 4/4 = 1 = s. \end{eqnarray}$

$\begin{eqnarray} \\ a&=&6,\ s=2 {\rm \ gives }\ b=3 \\ 6 + 3 &=&9,\ 6\times 3 = 18\ {\rm and }\ 18/9 = 2 = s. \end{eqnarray}$

$\begin{eqnarray} \\ a&=&12,\ s=3 {\rm \ gives }\ b=4 \\ 12 + 4 &=&16,\ 12\times 4 = 48\ {\rm and }\ 48/16 = 3 = s. \end{eqnarray}$

Davide Colli, age 13, Scuola Media Besozzi, who lives near Milan in Italy, showed that the number pair $a$ and $a(a - 1)$ is a solution for all values of $a$ (except $a = 1$). This is a special case of (1) with $s = a - 1$.

The Mount School York sent lots of examples taking the second number to be a multiple of the first. Charis Campbell, Christianne Eaves, Sheila Luk and Cheryl Wong proved that the pattern always works for two equal even numbers.

$$N + N = 2N, \ N\times N = N^2 \ {\rm and} \ N^2/2N = N/2$$

which is a whole number if and only if $N$ is even.

Sheila, Cheryl and Peach from year 10 gave the formula

$\begin{eqnarray} \\ a&=&A(k + 1),\ s = Ak, \ b=Ak(k + 1) \\ a + b &=& A(k + 1)^2, \ ab = A^2k(k + 1)^2 {\rm and }\ ab/(a + b) = Ak = s. \end{eqnarray}$

Again this is a special case of Farhan's result and it fits the pattern when $A$ and $k$ are whole numbers.

Farhan's formula given in (1) is not restricted to pairs of positive whole numbers as the following examples show:

$\begin{eqnarray} \\ a = 2, s=5 {\rm \ gives }\ b = (-10/3) \\ a + b = (-4/3),\ ab = (-20/3) \ {\rm and }\ ab/(a + b) = 5 = s. \end{eqnarray}$

$\begin{eqnarray} \\ a = 1/2, s=3 {\rm \ gives }\ b = (-3/5) \\ a + b = (-1/10),\ ab = (-3/10) \ {\rm and }\ ab/(a + b) = 3 = s. \end{eqnarray}$

Make a set of numbers that use all the digits from 1 to 9, once and once only. Add them up. The result is divisible by 9. Add each of the digits in the new number. What is their sum? Now try some other possibilities for yourself!