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

	Answered by 1549442205PVT last updated on 18/Sep/20
	$From the hypothesis abc^(−) ,bca^(−) ,cab^(−) form a G.P we infer abc^(−) ×cab^(−) =(bca^(−) )^2 ⇔((100a+10b+c)(100c+10a+b)= (100b+10c+a)^2 ⇔10000ac+1000bc+100c^2 +1000a^2 +100ab+10ac+100ab+10b^2 +bc =10000b^2 +100c^2 +a^2 +2000bc+200ab+20ac ⇔9990ac−999bc−9990b^2 +999a^2 =0 ⇔a^2 +10ac=10b^2 +bc ⇔(a+5c)^2 =25c^2 +10b^2 +bc ⇔100(a+5c)^2 =2500c^2 +100bc+1000b^2 ⇔(10a+50c)^2 =(50c+b)^2 +999b^2 ⇔(10a+100c+b)(10a−b)=999b^2 (2) =37.27.b^2 ⇒LHS⋮37 i)If b=9 then it is easy to see that 10a−b isn′t divisible 37.Hence 10a+100c+9⋮37.Since 10a+100c+9>111 and(10a+100c+9)=37k⇒37k−9⋮10 ⇒37k=999⇒10a−9=81⇒a=9 +)If 10.9+100c+9=999⇒c=9 We get abc^(−) =999 ii)If b=8 then 10a−8 isn′t divisible 37 10a+100c+8=37k⇒37k−8⋮10 this is impossible since 37k∈{259,629,999} By similar argument for the cases b=1,2,3,4,5,6,7 we see that ∄a,c satisfying(2) Thus,there is only one three−digits number abc^(−) =999 satisfying the condition of our problem$
	$From the hypothesis \overset{―}{abc}, \overset{―}{bca}, \overset{―}{cab}$ $form a G . P we infer \overset{―}{abc} \times \overset{―}{cab} = {(\overset{―}{bca})}^{2}$ $\Leftrightarrow ((100 a + 10 b + c) (100 c + 10 a + b) =$ ${(100 b + 10 c + a)}^{2}$ $\Leftrightarrow 10000 ac + 1000 bc + {100 c}^{2} + {1000 a}^{2}$ $+ 100 ab + 10 ac + 100 ab + {10 b}^{2} + bc$ $= {10000 b}^{2} + {100 c}^{2} + a^{2} + 2000 bc + 200 ab + 20 ac$ $\Leftrightarrow 9990 ac - 999 bc - {9990 b}^{2} + {999 a}^{2} = 0$ $\Leftrightarrow a^{2} + 10 ac = {10 b}^{2} + bc$ $\Leftrightarrow {(a + 5 c)}^{2} = {25 c}^{2} + {10 b}^{2} + bc$ $\Leftrightarrow 100 {(a + 5 c)}^{2} = {2500 c}^{2} + 100 bc + {1000 b}^{2}$ $\Leftrightarrow {(10 a + 50 c)}^{2} = {(50 c + b)}^{2} + {999 b}^{2}$ $\Leftrightarrow (10 a + 100 c + b) (10 a - b) = {999 b}^{2} (2)$ $= 37.27 . b^{2} \Rightarrow LHS ⋮ 37$ $i) If b = 9 then it is easy to see that$ $10 a - b {isn}^{'} t divisible 37 . Hence$ $10 a + 100 c + 9 ⋮ 37 . Since 10 a + 100 c + 9 > 111$ $and (10 a + 100 c + 9) = 37 k \Rightarrow 37 k - 9 ⋮ 10$ $\Rightarrow 37 k = 999 \Rightarrow 10 a - 9 = 81 \Rightarrow a = 9$ $+) If 10.9 + 100 c + 9 = 999 \Rightarrow c = 9$ $We get \overset{―}{abc} = 999$ $ii) If b = 8 then 10 a - 8 {isn}^{'} t divisible 37$ $10 a + 100 c + 8 = 37 k \Rightarrow 37 k - 8 ⋮ 10$ $this is impossible since 37 k \in {259, 629, 999}$ $By similar argument for the cases$ $b = 1, 2, 3, 4, 5, 6, 7 we see that ∄ a, c$ $satisfying (2)$ $Thus, there is only one three - digits$ $number \overset{―}{abc} = 999 satisfying the$ $condition of our problem$
	Commented by Rasheed.Sindhi last updated on 18/Sep/20

	$a, a r, {a r}^{2} . . . i s G P o n l y w h e n r \neq 1$ $. . .$
	Commented by mr W last updated on 18/Sep/20

	$111$ $222$ $333$ $. . . .$ $999$ $a r e o b v i o u s s o l u t i o n s .$
	Commented by 1549442205PVT last updated on 18/Sep/20

	$Thank Sir . I mistaked$
	Commented by 1549442205PVT last updated on 20/Sep/20

	$We assume that this is a special$ $geometry progression with q = 1$
	Commented by Rasheed.Sindhi last updated on 20/Sep/20

	$O k s i r!$
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