$2,3,5,6,7,10,11,12,13,...$ is the sequence of integers without all square and cube numbers. What is the $2019^{th}$ number?
By Using $PIE$
Answer:$2073$
Please give the full solution. How it is possible using PIE?
Re: BDMO 2019 : National : Junior : Pblm 05
Posted: Sat Mar 09, 2019 10:05 am
by samiul_samin
Take squares and cubes,
Remove the $6^{th}$ powers,
Giving us $54$ numbers to remove.Last removed number is $2025$
Thus we get our $2019^{th}$ number$=2019+54=2073$.
PIE=Inclusion Exclusion Principle
Try for smaller cases(70 or 100) to understand the solution clearly.
Take squares and cubes,
Remove the $6^{th}$ powers,
Giving us $54$ numbers to remove.Last removed number is $2025$
Thus we get our $2019^{th}$ number$=2019+54=2073$.
PIE=Inclusion Exclusion Principle
Try for smaller cases(70 or 100) to understand the solution clearly.
How do I identify that the last removed number is 2025?
Re: BDMO 2019 : National : Junior : Pblm 05
Posted: Sat Mar 09, 2019 9:02 pm
by samiul_samin
Try for smaller cases and write down the squares and cubes from $2000$ to $2100$.You will understand.I am sorry that my solution is not clear enough to understand.
Re: BDMO 2019 : National : Junior : Pblm 05
Posted: Sun Mar 10, 2019 11:44 am
by math_hunter
Where can I get a clear description about "PIE" and how to solve this problem using "PIE"?
Re: BDMO 2019 : National : Junior : Pblm 05
Posted: Sun Mar 10, 2019 12:55 pm
by vyper47
samiul_samin
So I assume it's basically this: we first remove all squares and cubes, and then add up the double-removed 6th powers
Well, samiul_samin's solution is perfectly correct. But for better understanding, I am elaborating this a bit. Okay, here is the full solution:
In order to find the $2019$th number of the series, we have to determine the perfect square and perfect cube numbers within $2019$ and the add that with 2019(Also have to do 2 more tasks then!). Okay, let's find the perfect squared numbers within $2019$ :
$$ \left \lfloor{\sqrt{2019}}\right \rfloor \ = 44$$
It means, there are 44 perfect squared numbers within 2019. Now, lets determine the perfect cubed numbers:
$$ \left \lfloor{\sqrt[3]{2019}}\right \rfloor \ = 12 $$
Then, there are 12 perfect cubed numbers within 2019. It means there are $44+12=56$ perfect square and perfect cube numbers within $2019$. Its a very general thought to add $56$ with $2019$ and say the answer is $56+2019=2075$. But as I mentioned before, we have to do 2 more tasks. The first one is to avoid repetitions. Avoiding repetitions mean we have to those numbers from $2075$, which are both perfect square and perfect cube numbers within $2019$ . Let, $x^2$ be the perfect squared numbers and $x^3$ be the perfect cubed numbers. Those, which are both perfect square and perfect cube numbers they are $x^6$. Let's find those $x^6$
$$\left \lfloor{\sqrt[6]{2019}}\right \rfloor \ = 3 $$. Then lets remove this $3$ numbers from $2075$ $2075-3=2072$. The last task is to find perfect square and perfect cube numbers within $2019$ and $2072$. You'll find that $45^2=2025$ is a perfect squared number within $2019$ and $2072$. No other numbers are perfect squared or perfect cubed within $2019$ and $2072$. Now, we have to add $1$ with $2072$ and thus, the answer is $2072+1+2073$. Then, the 2019th member of the series is 2073.!!
I guess you won't have any confusion regarding the problem after this.
