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Find the smallest integer solution to the equation 1/x^2 + 1/y^2 = 1/z^2

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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?

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Our Ages

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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.