Problem E and Solution

Missing the Fives III

Bobbi lists the positive integers, in order, excluding all multiples of \(5\). Her resulting list is \[1, 2, 3, 4, 6, 7, 8, 9, 11, 12, 13, 14, 16, 17, \ldots\] Determine the sum of the first \(2023\) integers in Bobbi’s list.

Note:

In solving this problem, it may be helpful to use the fact that the sum
of the first \(n\) positive integers is
equal to \(\tfrac{n(n+1)}{2}\). That
is, \[1 + 2 + 3 + \cdots + n =
\frac{n(n+1)}{2}\]

**Solution 1**

We begin by determining which integers are in Bobbi’s list. In each group of \(5\) consecutive integers beginning at \(1\), Bobbi lists \(4\) of the integers, since she leaves out each integer that is a multiple of \(5\). That is, in each of these groups of \(5\) integers, Bobbi’s list contains \(\frac{4}{5}\) of the integers.

Consider the positive integers from \(1\) to \(n\), where \(n\) is a multiple of \(5\). Of these \(n\) integers, Bobbi’s list contains \(\frac{4}{5}n\) integers. Since Bobbi’s list contains \(2023\) integers, which is not a multiple of \(4\), and \(2024\) is a multiple of \(4\), we will determine the sum of the first \(2024\) integers in Bobbi’s list and then subtract the \(2024\)th integer.

Now, \(\frac{4}{5}n= 2024\) or \(n = \frac{2024 \times 5}{4} = 2530\). So we need to determine the sum of the first \(2530\) positive integers with the integers that are multiples of \(5\) removed. That is, we need to determine the sum \[1 + 2 + 3 + 4 + 6 +\cdots+ 2524 + 2526 + 2527 + 2528 + 2529\] We will proceed to determine this sum by first calculating the sum of the integers from \(1\) to \(2530\). We will then subtract from that sum the sum of the integers in this list that are multiples of \(5\). We will also need to remove \(2529\), which is the \(2024\)th number in the list.

The sum of the integers from \(1\) to \(n\) is given by \(\frac{n(n+1)}{2}\), and so the sum of the integers from \(1\) to \(2530\) is equal to \(\frac{(2530)(2531)}{2} = 3\,201\,715\).

The sum of the multiples of \(5\) in
this list, \(5 + 10 + 15 + \cdots + 2520 +
2525 + 2530\), can be written as \(5(1
+ 2 + 3 +\cdots+ 504 + 505 + 506)\).

This is equal to \(5 \times
\frac{(506)(507)}{2} = 641\,355\).

Therefore, the sum of the first \(2023\) integers in Bobbi’s list is \(3\,201\,715 - 641\,355 - 2529= 2\,557\,831\).

**Solution 2**

In this solution, we will find the sum of the first \(2024\) integers in Bobbi’s list by pairing
up the integers, and then subtract the \(2024\)th integer in her list. From Solution
1, we know that the \(2024\)th number
in Bobbi’s list is \(2529\). Thus, the
sum of the first \(2024\) integers in
Bobbi’s list is \[1 + 2 + 3 + 4 + 6 + \cdots
+ 2524 + 2526 + 2527 + 2528 + 2529\] The sum of the first and
last integers in this list is \(1 + 2529 =
2530\).

The sum of the second integer and the second last integer is \(2 + 2528 = 2530\).

The sum of the third integer and the third last integer is \(3 + 2527 = 2530\).

We continue in this way moving toward the middle of the list. That is,
we move one number to the right of the previous first number, and one
number to the left of the previous second number. Doing so, we notice
that

when the first number in the new pair is one more than the previous first number, then the number it is paired with is one less than the previous second number, and

when the first number in the new pair is two more than the previous first number (as is the case when a multiple of \(5\) is omitted), then the number it is paired with is two less than the previous second number.

That is, as we continue moving toward the middle of Bobbi’s list, each pair will continue to have a sum equal to \(2530\). Since there are \(2024\) numbers in Bobbi’s list, there are \(1012\) such pairs, each having a sum of \(2530\). Thus, if Bobbi lists the positive integers, in order, leaving out the integers that are multiples of \(5\), the sum of the first \(2024\) integers in her list is \(1012 \times 2530 = 2\,560\,360\). However, this includes the \(2024\)th integer in her list. Therefore, the sum of the first \(2023\) integers in Bobbi’s list is \(2\,560\,360 - 2529 = 2\,557\,831\).