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Given a sequence $\{i\}_{i\in\mathbb{N}}$. A new sequence $\{p_n\}_{n\in\mathbb{N}}$ is obtained from $\{i\}_{i\in\mathbb{N}}$ by omitting all multiples of $3$ or $4$, but not $5$. Find $p_{2021}$.

I try as follows.

Let $S=\{1, 2, 3, ...,2021\}$.

Let $A=\{3, 6, 9, ..., 2019\}$, $B=\{4, 8, 12,..., 2020\}$, $C=\{60, 120, 180,..., 1980\}$, then

$|A|=673$, $|B|=505$. $|C|=33$

The number of omitted is $|S|-|A|-|B|+|C|=2021-673-505+33=876$ $p_{2021}=2021+|S|-|A|-|B|+|C|=2021+2021-673-505+33=2897$.

I'm very not sure with my answer. Anyone can help me?

Ongky Denny Wijaya
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    in additional 876 numbers, there will be some multiples of 3 and 4 which are to be omitted. (And the calculation of the number 876 is wrong too. for example the number 12 is considered twice) – dust05 Sep 15 '21 at 08:14
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    My suggestion is: the sequence is periodic, i.e. if $a$ is in the sequence then $a+60$ also is, where the number 60 is the least common multiple of 3, 4, 5. Justify this and try to use this fact – dust05 Sep 15 '21 at 08:16
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    You've tagged this as contest-math -- could you specify which contest and confirm that it's not ongoing please? – postmortes Sep 15 '21 at 08:49
  • This is only exercise for math contest. – Ongky Denny Wijaya Sep 15 '21 at 15:47

1 Answers1

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Let's count how many of the first $60n$ numbers satisfy these conditions.

Out of the first $60n$ numbers, there are $20n$ multiples of $3$; there are $15n$ multiples of $4$; there are $4n$ multiples of $15$; there are $3n$ multiples of $20$, and there are $n$ multiples of $60$.

Now consider the inequality $$60n-20n-15n+4n+5n+3n-n\geq 2021$$ $$36n\geq 2021$$ $$n>56.13888$$

$$60n>3368.33$$

This is also confirmed by the following simple Python code.

l=[]

for i in range(1,5000):
    l.append(i)


for i in range(1,5000):
    if (i%3==0 or i%4==0) and i%5!=0:
        l.remove(i)


print(l[2020])

$$p_{2021}=3370$$

You can run this code here.

F. A. Mala
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  • FYI You can edit your own solution if there is additional/different information that you want to showcase. Generally only add another solution is it's significantly different (EG Running code vs Counting argument) – Calvin Lin Sep 15 '21 at 16:28