Question Number 114290 by mathdave last updated on 18/Sep/20

Answered by 1549442205PVT last updated on 18/Sep/20

From the hypothesis abc^(−) ,bca^(−) ,cab^(−)   form a G.P we infer abc^(−) ×cab^(−) =(bca^(−) )^2   ⇔((100a+10b+c)(100c+10a+b)=  (100b+10c+a)^2   ⇔10000ac+1000bc+100c^2 +1000a^2   +100ab+10ac+100ab+10b^2 +bc  =10000b^2 +100c^2 +a^2 +2000bc+200ab+20ac  ⇔9990ac−999bc−9990b^2 +999a^2 =0  ⇔a^2 +10ac=10b^2 +bc  ⇔(a+5c)^2 =25c^2 +10b^2 +bc  ⇔100(a+5c)^2 =2500c^2 +100bc+1000b^2   ⇔(10a+50c)^2 =(50c+b)^2 +999b^2   ⇔(10a+100c+b)(10a−b)=999b^2 (2)  =37.27.b^2 ⇒LHS⋮37  i)If b=9 then it is easy to see that   10a−b isn′t divisible 37.Hence  10a+100c+9⋮37.Since 10a+100c+9>111  and(10a+100c+9)=37k⇒37k−9⋮10  ⇒37k=999⇒10a−9=81⇒a=9  +)If 10.9+100c+9=999⇒c=9  We get abc^(−) =999  ii)If b=8 then 10a−8 isn′t divisible 37  10a+100c+8=37k⇒37k−8⋮10  this is impossible since 37k∈{259,629,999}  By similar argument for the cases  b=1,2,3,4,5,6,7 we see that ∄a,c  satisfying(2)  Thus,there is only one three−digits    number abc^(−) =999 satisfying the  condition of our problem

Commented byRasheed.Sindhi last updated on 18/Sep/20

a,ar,ar^2 ...is GP  only when r≠1  ...

Commented bymr W last updated on 18/Sep/20

111  222  333  ....  999  are obvious solutions.

Commented by1549442205PVT last updated on 18/Sep/20

Thank Sir.I mistaked

Commented by1549442205PVT last updated on 20/Sep/20

We assume that this is a special  geometry progression with q=1

Commented byRasheed.Sindhi last updated on 20/Sep/20

Ok sir!