Try this beautiful problem from PRMO, 2019 based on Good numbers.

## Good numbers Problem | PRMO | Problem-12

A natural number \(k > \) is called good if there exist natural numbers

\(a_1 < a_2 < ………. < a_k\)

\(\frac{1}{\sqrt a_1} +\frac{1}{\sqrt a_2}+………………. +\frac{1}{\sqrt a_k}=1\)

Let \(f(n)\) be the sum of the first \(n\) good numbers, \(n \geq 1\). Find the sum of all values of \(n\) for which

\(f(n + 5)/f(n)\) is an integer.

- $20$
- $18$
- $13$

**Key Concepts**

Number theory

Good number

Integer

## Check the Answer

Answer:\(18\)

PRMO-2019, Problem 12

Pre College Mathematics

## Try with Hints

A number n is called a good number if It is a square free number.

Let \(a_1 ={A_1}^2\),\(a_2={A_2}^2\),………………\(a_k={A_k}^2\)

we have to check if it is possible for distinct natural number \(A_1, A_2………….A_k\) to satisfy,

\(\frac{1}{A_1}+\frac{1}{A_2}+………..+\frac{1}{A_k}=1\)

Can you now finish the problem ……….

For \(k = 2\); it is obvious that there do not exist distinct\( A_1, A_2\), such that \(\frac{1}{A_1}+\frac{1}{A_2}=1 \Rightarrow 2\) is not a good number

For \(k = 3\); we have \(\frac{1}{2} +\frac{1}{3}+\frac{1}{6}=1 \Rightarrow 3\) is a good number.

\(\frac{1}{2}+\frac{1}{2}\frac{1}{2}+\frac{1}{3}+\frac{1}{6}=1\) \(\Rightarrow 4\) is a good number

Let \(k\) wil be a good numbers for all \(k \geq 3\)

\(f(n) = 3 + 4 +… n\) terms =\(\frac{n(n + 5)}{2}\)

\(f(n + 5) =\frac{(n + 5)(n +10)}{2}\)

\(\frac{f(n+5}{f(n)}=\frac{n+10}{n}=1+\frac{10}{n}\)

Can you finish the problem……..

Therefore the integer for n = \(1\), \(2\), \(5\) and \(10\). so sum=\(1 + 2 + 5 + 10 = 18\).

## Other useful links

- https://www.cheenta.com/roots-of-cubic-equation-amc-10a-2010-problem-21/
- https://www.youtube.com/watch?v=uApjRLII6YI