Problem 510

Solution 1

In the list of integers beginning at 1, the 6$^{th}$ multiple of 5 is $6\times5=30$.

Thus, Tania has listed each of the integers from 1 to 29 with the exception of the positive multiples of 5 less than 30: $5,10,15,20,25$.

Therefore, just before Tania leaves out the $6^{th}$ multiple of 5, she has listed $29-5=24$ 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 Tania’s list of integers, she leaves out the integers that are multiples of 5, and so in every group of five integers, Tania lists four of these integers.

Thus, just before Tania leaves out the $6^{th}$ multiple of 5, she has listed $6\times4=24$ integers.
Solution 1

Tania writes 2019 just before leaving out 2020 (since 2020 is a multiple of 5).

Beginning at 1, 2020 is the 404$^{th}$ multiple of 5 since $\dfrac{2020}{5}=404$.

That is, the integers from 1 to 2020 contain 404 groups of 5 integers.

Each of these 404 groups contain one integer that is a multiple of 5, and so Tania leaves out 404 integers (including 2020) in the list of all integers from 1 to 2020.

If the $k^{th}$ integer in Tania’s list is 2019, then $k=2020-404=1616$.

Solution 2

Tania writes 2019 just before leaving out 2020 (since 2020 is a multiple of 5).

Beginning at 1, 2020 is the 404$^{th}$ multiple of 5 since $\dfrac{2020}{5}=404$.

That is, the integers from 1 to 2020 contain 404 groups of 5 integers.

In Tania’s list of integers, she leaves out the integers that are multiples of 5, and so in every group of five integers, Tania lists four of these integers.

If the $k^{th}$ integer in Tania’s list is 2019, then $k=404\times4=1616$.
Solution 1

We begin by determining which integers are in Tania’s list.

In each successive group of 5 consecutive integers beginning at 1, Tania 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, Tania’s list contains $\dfrac45$ of the integers.

Consider all positive integers from 1 to $n$, where $n$ is a multiple of 5.

Of these $n$ integers, Tania’s list contains $\dfrac45n$ integers.

Tania’s list contains 200 integers, and so $\dfrac45n=200$ or $n=\dfrac{200\times5}{4}=250$.

That is, if Tania lists the positive integers from 1 to 250 and leaves out the integers that are multiples of 5, her list will contain $\dfrac45\times 250=200$ integers.

We are required to determine the sum, $1+2+3+4+6+\cdots+244+246+247+248+249$, of the first 250 positive integers with the integers that are multiples of 5 removed.

We will proceed to determine this sum by first calculating the sum of all integers from 1 to 250, and then subtracting from that sum all integers in this list that are multiples of 5.

The sum of the integers from 1 to $n$ is given by $\dfrac12n(n+1)$, and so the sum of the integers from 1 to 250 is equal to $\dfrac12(250)(251)=31\,375$.

The multiples of 5 in this list, $5+10+15+\cdots+240+245+250$, can be written as $5(1+2+3+\cdots+48+49+50)$ by removing the common factor 5 (since each is a multiple of 5).

This sum is equal to $5\times \dfrac12(50)(51)=6375$.

If Tania lists the positive integers, in order, leaving out the integers that are multiples of 5, the sum of the first 200 integers in her list is $31\,375-6375=25\,000$.

Solution 2

As was shown in Solution 1, the sum of the first 200 integers in Tania’s list is the sum $1+2+3+4+6+\cdots+244+246+247+248+249$.

The sum of the first and last integers in this list is $1+249=250$.

The sum of the second integer and the second last integer is $2+248=250$.

The sum of the third integer and the third last integer is $3+247=250$.

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 recognize 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 Tania’s list, each pair will continue to have a sum equal to 250.

Since there are 200 numbers in Tania’s list, there are 100 such pairs, each having a sum equal to 250.

Thus, if Tania lists the positive integers, in order, leaving out the integers that are multiples of 5, the sum of the first 200 integers in her list is $250\times100=25\,000$.