# Problem of the Week Problem D and Solution Missing the Fives II

## Problem

Bobbi lists the positive integers, in order, excluding all multiples of 5. Her resulting list is $1, 2, 3, 4, 6, 7, 8, 9, 11, 12, 13, 14, 16, 17, \ldots$ If the $$n$$th integer in Bobbi’s list is $$2023$$, what is the value of $$n$$?

## Solution

Solution 1

Note that $$2023$$ is two integers before $$2025$$, which is a multiple of $$5$$.
Beginning at $$1$$, $$2025$$ is the $$405$$th multiple of $$5$$, since $$\frac{2025}{5} = 405$$. That is, the integers from $$1$$ to $$2025$$ contain $$405$$ groups of $$5$$ integers.
Each of these $$405$$ groups contain one integer that is a multiple of $$5$$, and so Bobbi leaves out $$406$$ integers (including $$2024$$) in the list of all integers from $$1$$ to $$2025$$. If the $$n$$th integer in Bobbi’s list is $$2023$$, then $$n = 2025 - 406 = 1619$$.

Solution 2

Note that $$2023$$ is two integers before $$2025$$, which is a multiple of $$5$$.
Beginning at $$1$$, $$2025$$ is the $$405$$th multiple of $$5$$, since $$\frac{2025}{5} = 405$$. That is, the integers from $$1$$ to $$2025$$ contain $$405$$ groups of $$5$$ integers.
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. However, she also does not list $$2024$$. Thus, if the $$n$$th integer in Bobbi’s list is $$2023$$, then $$n = 405 \times 4 - 1= 1619$$.