National Junior 2020 Problem

For students of class 6-8 (age 12 to 14)
Shahriar Amin Arif
Posts:2
Joined:Sat Dec 26, 2020 5:54 pm
National Junior 2020 Problem

Unread post by Shahriar Amin Arif » Sun Apr 04, 2021 10:57 pm

Plz solve this problem
Attachments
lllll.png
lllll.png (76.69KiB)Viewed 8925 times

User avatar
sakib17442
Posts:15
Joined:Sun Apr 11, 2021 10:04 pm
Contact:

Re: National Junior 2020 Problem

Unread post by sakib17442 » Wed Apr 21, 2021 7:46 pm

Every number written in this solution is in base-$10$.
We need to find $n$ such that $n=(d_kd_{k-1}\dots d_1d_0)_7$ and $2n=(d_kd_{k-1}\dots d_1d_0)_{10}$ where $d_i$ s are digits in the base-$7$ representation of $n$. So, for each $i$, $0\leq d_i\leq 6$.
We can write the whole thing with this equation:
$\sum_{i=0}^{k}10^id_i=2\sum_{i=0}^{k}7^id_i\Leftrightarrow\sum_{i=0}^{k}(10^i-2\cdot7^i)d_i=0$.
Let's call $2n$ a good number if $n$ has the desired properties given in this question.

Lemma-1 : For integer $m\geq3$, $10^m-2\cdot7^m>30$
Proof:
Base case : For $m=3$, $10^m-2\cdot7^m=314>30$, so our lemma is true for $m=3$.
Induction hypothesis : Let's assume the lemma is true for $m=j$.
Inductive step : For $m=j+1$,
$10^{j+1}-2\cdot7^{j+1}=3\cdot10^j+7(10^j-2\cdot7^j)>30$
So, our lemma is true for $m=j+1$, that completes our proof of this lemma using induction.

Claim-1 : There is no good number with more than $3$ digits in base-$10$
For this, we will assume that $k\geq3$ and prove that the minimum value of $S_k=\sum_{i=0}^{k}(10^i-2\cdot7^i)d_i>0$.
Proof :
To achieve the minimum value of $S_k$, we need $d_k=1$, since the most significant digit or the first digit is nonzero and as small as possible.
We also need $d_i=0$ when $i\neq k$ and $10^i-2\cdot7^i>0$ and $d_i=6$ when $10^i-2\cdot7^i<0$.
That means we will minimize the value of $d_i$ when we are adding a non-negative term and maximize it when we are adding a negative term to make sure we are adding the minimum possible and subtracting the maximum possible to get the minimum value of $S_k$.
But $10^i-2\cdot7^i<0$ only when $i\in\{0,1\}$. For $i=2$, $10^i-2\cdot7^i=2$ and the case $i>2$ has been explored in lemma-$1$
So, the minimum value of $S_k=(10^k-2\cdot7^k)\cdot1+(10^1-2\cdot7^1)\cdot6+(10^0-2\cdot7^0)\cdot6=(10^k-2\cdot7^k)-30$
By lemma-$1$, $S_k\geq min(S_k)=(10^k-2\cdot7^k)-30>0$ for $k\geq3$.
But $S_k=0$ is a neccessary condition for such a good number to exist and we just proved that this is not possible for $k\geq3$. Or, equivalently, there is no good number with more than $3$ digits. So, claim-$1$ is pretty much proved.

So, now we can finally and peachfully give all of our attention to $k=2$ case. :P :P
$S_2=2d_2-4d_1-d_0=0$
To get the largest possible number, let's take our most significant digit $d_2=6$ [Remember that $d_i$ is a digit in base $7$, so $0\leq d_i\leq 6$]
So, our equation becomes:
$12=4d_1+d_0$
Similarly, maximum possible value of $d_1$ is $3$ and this forces $d_0=0$.
That means, the largest good number $2n=630\Leftrightarrow n=315$.
It is easy to check that $n=315$ has the desired properties. [Actually it's already proven]

I just copied and pasted The solution of Anindya Biswas. Check out the actual solution by Anindya Biswas . Here is the link: viewtopic.php?f=13&t=6103
Games You can't win because you'll play against yourself.
---Dr.Seuss

pasterzop
Posts:2
Joined:Wed Sep 28, 2022 7:01 pm

Re: National Junior 2020 Problem

Unread post by pasterzop » Sat Oct 01, 2022 5:05 pm

that's an interesting question

pasterzop
Posts:2
Joined:Wed Sep 28, 2022 7:01 pm

Re: National Junior 2020 Problem

Unread post by pasterzop » Mon Oct 03, 2022 5:08 pm

i tried solving the equation but it didn't worked tellculvers com survey taco bell breakfast hours

Post Reply