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Problem of the Week
Problem C and Solution
Missing the Fives I

Problem

Bobbi lists the positive integers, in order, excluding all multiples of \(5\). Her resulting list is \[1, 2, 3, 4, 6, 7, 8, 9, 11, 12, 13, 14, 16, 17, \ldots\] How many integers has Bobbi listed just before she leaves out the \(2023\)rd multiple of \(5\)?

Solution

Solution 1

In the list of integers beginning at \(1\), the \(2023\)th multiple of \(5\) is \(2023 \times 5 = 10\,115\). Thus, Bobbi has listed each of the integers from \(1\) to \(10\,114\) with the exception of the positive multiples of \(5\) less than \(10\,115\). Since \(10\,115\) is the \(2023\)rd multiple of \(5\), Bobbi will not write \(2022\) multiples of \(5\).

Therefore, just before Bobbi leaves out the \(2023\)rd multiple of \(5\), she has listed \(10\,114 - 2022 = 8092\) integers.

Solution 2

Beginning at \(1\), each group of five integers has one integer that is a multiple of \(5\). For example, the first group of five integers, \(1\), \(2\), \(3\), \(4\), \(5\), has one multiple of \(5\) (namely \(5\)), and the second group of five integers, \(6\), \(7\), \(8\), \(9\), \(10\), has one multiple of \(5\) (namely \(10\)). In Bobbi’s list of integers, she leaves out the integers that are multiples of \(5\), and so in every group of five integers, Bobbi lists four of these integers. Thus, just before Bobbi leaves out the \(2023\)rd multiple of \(5\), there were \(2023\) of these groups. Therefore, she has listed \(2023 \times 4 = 8092\) integers.