Sudipta noticed that his birth year $1978$
is a memorable year in which two number of two sides are $19$ and $78$. The sum of this two number is $97$ which is it's middle number. Now it is $2018$. How many years will Sudipta wait for the next memorable year?
Source : Gonitzoggo, Link: https://gonitzoggo.com/archive/problem/178
Memorable Year
- sakib17442
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- Anindya Biswas
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Re: Memorable Year
Let's assume $1000a+100b+10c+d$ is a memorable year where $a,b,c,d\in\mathbb{Z}; 2\leq a\leq9; 0\leq b,c,d\leq9$. [That means they are valid digits in base $10$.]sakib17442 wrote: ↑Mon Jul 12, 2021 12:44 amSudipta noticed that his birth year $1978$
is a memorable year in which two number of two sides are $19$ and $78$. The sum of this two number is $97$ which is it's middle number. Now it is $2018$. How many years will Sudipta wait for the next memorable year?
Source : Gonitzoggo, Link: https://gonitzoggo.com/archive/problem/178
They must satisfy the equation,
$10a+b+10c+d=10b+c$
$\Longleftrightarrow 10a+9c+d=9b$
To get the minimum such year after $2018$, let's guess $a=2$.
So, $20+d=9(b-c)$.
Since $9\mid 20+d$ and $20\leq 20+d\leq 29$,
We must have, $20+d=27\Longleftrightarrow d=7$
From here, we conclude, $b-c=3$. To minimize the value of $b$, let's use the solution $b=3, c=0$.
From here, we conclude that $2307$ is the closest memorable year after $2018$.
So, Sudipta should wait about $\boxed{2307-2018=289}$ years only.
Hoping that people remember this post after this many years and celebrates with great joy
"If people do not believe that mathematics is simple, it is only because they do not realize how complicated life is."
— John von Neumann
— John von Neumann