BdMO National 2021 Junior Problem 7

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Anindya Biswas
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BdMO National 2021 Junior Problem 7

Unread post by Anindya Biswas » Mon Apr 12, 2021 11:54 am

একটা সংখ্যাকে বিলম্বী-কিশোর বলা হবে যদি সেটা তার অঙ্কগুলোর যোগফলের \(19\) গুণ হয়। কতগুলো বিলম্বী-কিশোর সংখ্যা আছে?

A late-teen number is a positive integer which is $19$ times the sum of its own digits. Determine how many late-teen numbers are there.
"If people do not believe that mathematics is simple, it is only because they do not realize how complicated life is."
John von Neumann

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Anindya Biswas
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Re: BdMO National 2021 Junior Problem 7

Unread post by Anindya Biswas » Wed Aug 25, 2021 3:49 am

Anindya Biswas wrote:
Mon Apr 12, 2021 11:54 am
একটা সংখ্যাকে বিলম্বী-কিশোর বলা হবে যদি সেটা তার অঙ্কগুলোর যোগফলের \(19\) গুণ হয়। কতগুলো বিলম্বী-কিশোর সংখ্যা আছে?

A late-teen number is a positive integer which is $19$ times the sum of its own digits. Determine how many late-teen numbers are there.
Detailed solution :
Let's assume, $N$ is a late-teen number. $a_0,a_1,a_2,\dots,a_n$ are it's digit such that,
$N=a_0+10a_1+10^2a_2+\dots+10^na_n\cdots \cdots \cdots(1)$
Where $0\leq a_i\leq9$ for all $0\leq i\leq n$ and $a_n\geq1$ when $n\geq1$.

By definition, $N$ is late-teen.
$\therefore N=19\left(a_0+a_1+a_2+\dots+a_n\right)\cdots\cdots\cdots(2)$
It's now a Diophantine equation of $n+1$ variables. What do we do in such situations? We use inequalities to reduce cases.

Before using any inequality, let me ask you, which of this expression should be bigger in general sense? Intuitively, the factor $10^n$ dominates for large $n$ in $(1)$, while in $(2)$, there is just a factor of $19$ with $a_n$. So, keeping $(1)$ at the greater side of the inequality shouldn't give any interesting cases. Intuitively, we can get interesting results by keeping results from $(1)$ at the lesser side of the inequality.

From $(1)$,
$N\geq 10^n$
From $(2)$,
$N\leq19\underbrace{(9+9+9+\dots+9)}_{n+1}=171(n+1)$
We can combine this results to get,
$10^n\leq N\leq 171(n+1)$
$\Rightarrow 10^n < 171(n+1)$ [Equality never holds since $171\nmid10^n$]
We can check that this inequality is only true when $n\in\{0,1,2\}$. Since this is a necessary condition for $N$ to be a late-teen integer, our case reduced to only at most $3$ variables. We will handle it case-by-case.
  • Case 0 : $n=0$ :
    $\therefore N=a_0=19a_0$
    $\Rightarrow a_0=0$
    $\therefore N=0$.
    So, we've got $1$ solution.
    [The phrasing of the question is confusing to understand if BdMO considered this as valid or not. They should have specified what a "number" is, is it complex? A fraction? A positive integer? Specially in a situation where we can't ask for clarification or show our works, a poorly defined question is not suitable.]
  • Case 1 : $n=1$ :
    $\therefore N=a_0+10a_1=19a_0+19a_1$
    $\Rightarrow 18a_0+9a_1=0$
    $\Rightarrow 2a_0+a_1=0$
    $\Rightarrow a_0=a_1=0$.
    But since $a_1\geq1$, we've arrived at a contradiction. So, this case has no valid solution.
  • Case 2 : $n=2$ :
    $\therefore N=a_0+10a_1+100a_2=19a_0+19a_1+19a_2$
    $\Rightarrow 81a_2=9a_1+18a_0$
    $\Rightarrow 9a_2=a_1+2a_0$
    $\Rightarrow 9a_2\leq 9+2\cdot9$
    $\Rightarrow a_2\leq 3$

    When $a_2=1$,
    The value of $a_0$ can be $0,1,2,3,4$.

    When $a_0=2$,
    The value of $a_0$ can be $5,6,7,8,9$.

    When $a_0=3$,
    The value of $a_0$ can only be $9$.

    So, there are total $11$ solutions for this case.
Considering this $3$ cases, we conclude that there are total $\boxed{1+0+11=12}$ non-negative late-teen numbers.
Here's the complete list : $0, 114, 133, 152, 171, 190, 209, 228, 247, 266, 285, 399$
"If people do not believe that mathematics is simple, it is only because they do not realize how complicated life is."
John von Neumann

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