# 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.