একটা ধনাত্মক পূর্ণসংখ্যা \(n\)-কে মনোরম বলা হবে যদি এটার কমপক্ষে \(3\)-টা প্রকৃত উৎপাদক থাকে এবং এটা তার সবচেয়ে বড় তিনটা প্রকৃত উৎপাদকের যোগফলের সমান হয়। যেমন \(6\) একটা মনোরম সংখ্যা কারণ \(6\)-এর সবচেয়ে বড় তিনটা প্রকৃত উৎপাদক হলো \(3\), \(2\), \(1\) এবং \(6=3+2+1\)। \(6000\)-এর চেয়ে বড় না, এমন কতগুলো মনোরম সংখ্যা আছে?

A positive integer $n$ is called nice if it has at least $3$ proper divisors and it is equal to the sum of its three largest proper divisors. For example, $6$ is nice because its largest proper divisors are $3,2,1$ and $6=3+2+1$. Find the number of nice integers not greater than $6000$.

## BdMO National 2021 Higher Secondary Problem 9

- Anindya Biswas
**Posts:**247**Joined:**Fri Oct 02, 2020 8:51 pm**Location:**Magura, Bangladesh-
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### BdMO National 2021 Higher Secondary Problem 9

"If people do not believe that mathematics is simple, it is only because they do not realize how complicated life is."

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**John von Neumann**- Anindya Biswas
**Posts:**247**Joined:**Fri Oct 02, 2020 8:51 pm**Location:**Magura, Bangladesh-
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### Solution :

Let me start with an exercise. Find three distinct positive integers $a,b,c$ where $a<b<c$ such that \[\frac1a+\frac1b+\frac1c=1\] You may post the solution below.

Well, the only solution to this equation is $(a,b,c)=(2,3,6)$.

Let's assume, $n$ is a nice integer. Let $\frac{n}{x},\frac{n}{y},\frac{n}{z}$ be the largest three proper divisor of $n$ where $\frac{n}{x}>\frac{n}{y}>\frac{n}{z}$. Then we got,

$n=\frac{n}{x}+\frac{n}{y}+\frac{n}{z}\Longleftrightarrow \frac{1}{x}+\frac{1}{y}+\frac{1}{z}=1$

$\therefore x=2,y=3,z=6$

So, $n$ is divisible by $2,3,6$. Also, $n$ is not divisible by $4,5$ since otherwise $\frac{n}{4}$ or $\frac{n}{5}$ would be one of the largest three proper divisors which is not the case. So, we can consider this system of linear congruences to solve for $n$:

\[\begin{array} \ n\equiv0\pmod3 \\ n\equiv2\pmod{4} \\ n\equiv r\pmod{5} \end{array}\] Where $r\in\{1,2,3,4\}$

By Chinese Remainder Theorem, this congruences has unique solutions modulo $60$. So, for each value of $r$, we get a unique number between $1$ and $60$ which is nice. So, there are $4$ nice integers less than or equal to $60$.

So, there are $\boxed{400}$ nice integers not greater than $6000$.

Well, the only solution to this equation is $(a,b,c)=(2,3,6)$.

Let's assume, $n$ is a nice integer. Let $\frac{n}{x},\frac{n}{y},\frac{n}{z}$ be the largest three proper divisor of $n$ where $\frac{n}{x}>\frac{n}{y}>\frac{n}{z}$. Then we got,

$n=\frac{n}{x}+\frac{n}{y}+\frac{n}{z}\Longleftrightarrow \frac{1}{x}+\frac{1}{y}+\frac{1}{z}=1$

$\therefore x=2,y=3,z=6$

So, $n$ is divisible by $2,3,6$. Also, $n$ is not divisible by $4,5$ since otherwise $\frac{n}{4}$ or $\frac{n}{5}$ would be one of the largest three proper divisors which is not the case. So, we can consider this system of linear congruences to solve for $n$:

\[\begin{array} \ n\equiv0\pmod3 \\ n\equiv2\pmod{4} \\ n\equiv r\pmod{5} \end{array}\] Where $r\in\{1,2,3,4\}$

By Chinese Remainder Theorem, this congruences has unique solutions modulo $60$. So, for each value of $r$, we get a unique number between $1$ and $60$ which is nice. So, there are $4$ nice integers less than or equal to $60$.

So, there are $\boxed{400}$ nice integers not greater than $6000$.

"If people do not believe that mathematics is simple, it is only because they do not realize how complicated life is."

—

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**John von Neumann**### Re: BdMO National 2021 Higher Secondary Problem 9

This also came as Secondary Problem 10.Anindya Biswas wrote: ↑Sun Apr 11, 2021 10:31 amA positive integer $n$ is called nice if it has at least $3$ proper divisors and it is equal to the sum of its three largest proper divisors. For example, $6$ is nice because its largest proper divisors are $3,2,1$ and $6=3+2+1$. Find the number of nice integers not greater than $6000$.

"When you change the way you look at things, the things you look at change." - Max Planck

### Re: Solution :

I just realised somethingAnindya Biswas wrote: ↑Sun Apr 11, 2021 11:11 amLet me start with an exercise. Find three distinct positive integers $a,b,c$ where $a<b<c$ such that \[\frac1a+\frac1b+\frac1c=1\] You may post the solution below.

Well, the only solution to this equation is $(a,b,c)=(2,3,6)$.

Let's assume, $n$ is a nice integer. Let $\frac{n}{x},\frac{n}{y},\frac{n}{z}$ be the largest three proper divisor of $n$ where $\frac{n}{x}>\frac{n}{y}>\frac{n}{z}$. Then we got,

$n=\frac{n}{x}+\frac{n}{y}+\frac{n}{z}\Longleftrightarrow \frac{1}{x}+\frac{1}{y}+\frac{1}{z}=1$

$\therefore x=2,y=3,z=6$

So, $n$ is divisible by $2,3,6$. Also, $n$ is not divisible by $4,5$ since otherwise $\frac{n}{4}$ or $\frac{n}{5}$ would be one of the largest three proper divisors which is not the case. So, we can consider this system of linear congruences to solve for $n$:

\[\begin{array} \ n\equiv0\pmod3 \\ n\equiv2\pmod{4} \\ n\equiv r\pmod{5} \end{array}\] Where $r\in\{1,2,3,4\}$

By Chinese Remainder Theorem, this congruences has unique solutions modulo $60$. So, for each value of $r$, we get a unique number between $1$ and $60$ which is nice. So, there are $4$ nice integers less than or equal to $60$.

So, there are $\boxed{400}$ nice integers not greater than $6000$.

I did a similar solution like this...

Every nice integer is of this format-

6*k where k is not a multiple of 2 and 5

Which results there is 400 nice integers inside 6000 range

But unfortunately I submitted the values of 'k' for which '6k' is "not" a nice integer.I gave 600 as the answer....now I realised I absolutely forgot to substract it from 1000(there are 6000/6=1000 total values of k) to get the real answer :"(

Online national cost me a P10(Secondary)