Problem
of the Week
Problem
E and Solution
Favourite
Numbers

Problem

Adrian likes all the numbers, but some are his favourites. He created
a flowchart to help people determine whether or not a given five-digit
positive integer is one of his favourites.

Start with a five-digit positive integer and proceed through the following steps:

Does the integer contain any even digits or repeated digits?

If YES, then the integer is not a
favourite.

If NO, then go to 2.

Is the thousands digit larger than both the hundreds digit and
the ten thousands digit?

If NO, then the integer is not a
favourite.

If YES, then go to 3.

Is the tens digit larger than both the hundreds digit and the
units digit?

If NO, then the integer is not a
favourite.

If YES, then the integer is a favourite.

How many favourite five-digit positive integers does Adrian have?

Solution

We write one of Adrianâ€™s favourite five-digit positive integers as
\(V W X Y Z\), where each letter
represents a digit.

Since this integer does not contain any even or repeated digits, then
it is created using the digits \(1\),
\(3\), \(5\), \(7\), and \(9\), in some order. We want to count the
number of ways of assigning \(1\),
\(3\), \(5\), \(7\), \(9\)
to the digits \(V\), \(W\), \(X\), \(Y\), \(Z\)
so that the answers to the second two questions in the flowchart are
both yes.

Since the thousands digit is larger than both the hundreds digit and
the ten thousands digit, then \(W >
X\) and \(W > V\). Since the
tens digit is larger than both the hundreds digit and the units digit,
then \(Y > X\) and \(Y > Z\).

The digits \(1\) and \(3\) cannot be placed as \(W\) or \(Y\), since \(W\) and \(Y\) are larger than both of their
neighbouring digits, while \(1\) is
smaller than all of the other digits and \(3\) is larger than only one of the other
possible digits.

The digit \(9\) cannot be placed as
\(V\), \(X\), or \(Z\) since it is the largest possible digit
and so cannot be smaller than \(W\) or
\(Y\) . Thus, \(9\) must be placed as \(W\) or as \(Y\). Therefore, the digits \(W\) and \(Y\) are \(9\) and either \(5\) or \(7\).

Suppose that \(W = 9\) and \(Y = 5\). The number is thus \(V 9 X 5 Z\). Neither \(X\) or \(Z\) can equal \(7\) since \(7
> 5\), so \(V = 7\). It
follows that \(X\) and \(Z\) are \(1\) and \(3\), or \(3\) and \(1\). There are \(2\) possible integers in this case.
Similarly, if \(Y = 9\) and \(W = 5\), there are \(2\) possible integers.

Suppose that \(W = 9\) and \(Y = 7\). The number is thus \(V 9 X 7 Z\). The digits \(1\), \(3\), and \(5\) can then be placed in any of the
remaining spots. There are \(3\)
choices for the digit \(V\). For each
of these choices, there are \(2\)
choices for \(X\), and then \(1\) choice for \(Z\). There are thus \(3 \times 2 \times 1 = 6\) possible integers
in this case. Similarly, if \(Y = 9\)
and \(W = 7\), there are \(6\) possible integers.

Therefore, Adrian has \(2 + 2 + 6 + 6 =
16\) favourite positive five-digit integers.