Question

Call a natural number $n$ faithful, if there exist numbers $a<b<c$ such that $a$ divides $b,b$ divides $c$ and $n=a+b+c$. Show that all but a finite number of natural numbers are faithful. Find the sum of all natural numbers which are not faithful.

Answer 1

Suppose $n\in N$ is faithful.
Let $k\in N$ and consider $kn$. Since $n=a+b+c$, with $a>b>c,c|bandb|a$, we see that $kn=ka+kb+kc$ which shows that $kn$ is faithful.
Let $p>5$ be a prime.
Then $p$ is odd and $p=(p-3)+2+1$ shows that $p$ is faithful.
If $n\in N$ contains a prime factor $p>5$, then the above observation shows that $n$ is faithful.
This shows that a number which is not faithful must be of the form $2^{\alpha }{3}^{\beta }{5}^{\gamma }$.
We also observe that $2^4=16=12+3+1,3^2=9=6+2+1and{5}^2=25=22+2+1$, so that $2^4,3^{2}and{5}^2$ are faithful.
$\therefore ninN$ is also faithful if it contains a factor of the form $2^{\alpha }$ where $\alpha \ge 4$; a factor of the form $3^{\beta }$ where $\beta \ge 2$; or a factor of the form $5^{\gamma }$ where $\gamma \ge 2$.
Thus the numbers which are not faithful are of the form $2^{\alpha }{3}^{\beta }{5}^{\gamma }$, where $\alpha \le 3,\beta \le 1and\gamma \le 1$. We may enumerate all such numbers : $1,2,3,4,5,6,8,10,12,15,20,24,30,40,60,120$.
Among these $120=112+7+1,60=48+8+4,40=36+3+1,30=18+9+3,20=12+6+2,15=12+2+1,and10=6+3+1$.
It is easy to check that the other numbers cannot be written in the required form. Hence the only numbers which are not faithful are $1,2,3,4,5,6,8,12,24$.
Their sum is $65$.