Skip over navigation
To find two positive integers adding up to 100, one a multiple of 7 and the other a multiple of 11, you can try 7 + 93, 14 + 86, 21 + 79, 28 + 72, 35 + 65, 42 + 58, 49 + 51 and 56 + 44 ... etc. and the only pair satisfying this condition is 56 and 44. Hence $x=8$ and $y=4$ is one particular solution of $7x + 11y =100$ and (it turns out) the only solution where both $x$ and $y$ are positive integers.
Jonathan Gill of St Peters' College, Adelaide restricted his search for solutions of this equation to positive integers and found the solution $x=8$ and $y=4$ by trying values of $x$ from $x=1$ to $x=13$ in turn.
Andaleeb Ahmed, age 17, Woodhouse Sixth Form College, London extended the search for solutions to include negative integers. This is Andaleeb's solution: The equation $7x + 11y = 100$ has solutions:
$\begin{eqnarray} \\ x_1 &=& 8 & \ \ \ & y_1 &=& 4\\ x_2 &=& -3 & \ \ \ & y_2 &=& 11\\ x_3 &=& -14 & \ \ \ & y_3 &=& 18\\ x_4 &=& -25 & \ \ \ & y_4 &=& 25\\ \end{eqnarray}$
From this we notice that the difference between two consecutive $x$ terms is -11 and the difference in $7x$ is therefore -77. The difference between two consecutive $y$ terms is +7 and so the difference in $11y$ is +77. This enables us to deduce the general terms $x_n$ and $y_n$ which are:
$\begin{eqnarray} \\ x_n &=& 8 - 11(n-1)\\ y_n &=& 4 + 7(n-1). \end{eqnarray}$
Thus by substituting any values of $n$ (it can be any integer including 0 and negative integers) in these expressions we can find infinitely many integer solutions of $7x + 11y = 100$.
In general if we can find one particular solution to an equation of this type we can use this method to find an infinite set of solutions.
In Particular
Age 14 to 16
Challenge Level 
To find two positive integers adding up to 100, one a multiple of 7 and the other a multiple of 11, you can try 7 + 93, 14 + 86, 21 + 79, 28 + 72, 35 + 65, 42 + 58, 49 + 51 and 56 + 44 ... etc. and the only pair satisfying this condition is 56 and 44. Hence $x=8$ and $y=4$ is one particular solution of $7x + 11y =100$ and (it turns out) the only solution where both $x$ and $y$ are positive integers.
Jonathan Gill of St Peters' College, Adelaide restricted his search for solutions of this equation to positive integers and found the solution $x=8$ and $y=4$ by trying values of $x$ from $x=1$ to $x=13$ in turn.
Andaleeb Ahmed, age 17, Woodhouse Sixth Form College, London extended the search for solutions to include negative integers. This is Andaleeb's solution: The equation $7x + 11y = 100$ has solutions:
$\begin{eqnarray} \\ x_1 &=& 8 & \ \ \ & y_1 &=& 4\\ x_2 &=& -3 & \ \ \ & y_2 &=& 11\\ x_3 &=& -14 & \ \ \ & y_3 &=& 18\\ x_4 &=& -25 & \ \ \ & y_4 &=& 25\\ \end{eqnarray}$
From this we notice that the difference between two consecutive $x$ terms is -11 and the difference in $7x$ is therefore -77. The difference between two consecutive $y$ terms is +7 and so the difference in $11y$ is +77. This enables us to deduce the general terms $x_n$ and $y_n$ which are:
$\begin{eqnarray} \\ x_n &=& 8 - 11(n-1)\\ y_n &=& 4 + 7(n-1). \end{eqnarray}$
Thus by substituting any values of $n$ (it can be any integer including 0 and negative integers) in these expressions we can find infinitely many integer solutions of $7x + 11y = 100$.
In general if we can find one particular solution to an equation of this type we can use this method to find an infinite set of solutions.
You may also like
Rudolff's Problem
A group of 20 people pay a total of £20 to see an exhibition. The admission price is £3 for men, £2 for women and 50p for children. How many men, women and children are there in the group?
