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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.

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Solution

## Suppose n∈N is faithful. Let k∈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∈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 2α3β5γ. We also observe that 24=16=12+3+1,32=9=6+2+1and52=25=22+2+1, so that 24,32and52 are faithful. ∴ninN is also faithful if it contains a factor of the form 2α where α≥4; a factor of the form 3β where β≥2; or a factor of the form 5γ where γ≥2. Thus the numbers which are not faithful are of the form 2α3β5γ, where α≤3,β≤1andγ≤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.

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