common ratio can be 2 (a has to be 2 in this case) or 3 (a has to be 1 in this case). then case work. most probably no solution.

EDIT: as pointed out by me in my 2nd post, this solution is wrong

yes $a=b=c$ is the only solution.

no non-trivial stuff.. common ratio = 1 yields the obvious solns..

Guys.. one of my friends pointed out that only abc is given as a 3 digit no... so infact there are 18 solutions everyone just got trolled

Quote: EDIT: as pointed out by me in my 2nd post, this solution is wrong what's wrong?

TheOneYouWant wrote: Guys.. one of my friends pointed out that only abc is given as a 3 digit no... so infact there are 18 solutions everyone just got trolled i dont think its wrong. i took the numbers to be 3 digit only.

I got 11 solutions for this, not sure where the other 7 are...TheOneYouWant can you say the 7 solutions I am missing please?

@YESmaths can u give all the questions of this paper ?

I mean RMO Mumbai region's paper

What i meant before was that it is possible that $(bca)$ and $(cab)$ are not 3 digit numbers, but $(abc)$ is, which leads to other solutions. But i don't think the examiners would cut marks for this anyways

What if the common ratio isn't an integer? I don't think it's necessary. Anyone with a proper solution?

We have given three natural numbers $x,y,z$ with three digits satisfying $(1) \;\; x = 100a + 10b + c$, $(2) \;\; y = 100b + 10c + a$, $(3) \;\; z = 100c + 10a + b$, $(4) \;\; xz = y^2$. According to equation (4) there are three natural numbers $k,u,v$, where $k$ is squarefree, s.t. $(5) \;\; (x,y,z) = (ku^2,kuv,kv^2)$. Combining (1)-(3) and (5) we obtain $(6) \;\; 999a = k(10u^2 - v^2)$, $(7) \;\; 999b = kv(10v - u)$, $(8) \;\; 999c = ku(10v - u)$. Let consider the following two cases: Case 1: $u=v$. Then $a=b=c$ by (6)-(8), i.e. $(a,b,c) = (m,m,m)$, where $1 \leq m \leq 9$, are the solutions of the system of equations (1)-(4). Case 2: $u \neq v$. Assume $37 | k$. According to (5) $999 = 27 \cdot 37 \geq x = ku^2 \geq 37v^2$ implies $u^2 \leq 27$, i.e. $u \leq 5$. Likewise $v \leq 5$ by (5), meaning $(9) \;\; 1 \leq u,v \leq 5$. The fact that $k$ is squarefree combined with equation (7) yields $(10) \;\; 9 \mid v(v - u)$. Seeing that $1 \leq |v-u| \leq 4$ by (9) and the fact that $u \neq v$, we obtain $3|v$ and $9 | uv$ by (10), yielding $3|u$ by (9). Consequently $u=v=3$ by (9), which is impossible since $u \neq v$. Hence $37 \nmid k$ by contradiction. The fact that $1 \leq u,v \leq 31 < 37$ (since $32^2 > 1000 > x=ku^2 \geq u^2$ and $32^2 > 1000 > z = kv^2 \geq v^2$) combined with (7) or (8) give us $(11) \;\; 37 \mid 10v - u$. Thus according to (11) there exists a non-negative integer $t$ s.t. $(12) \;\; 10v - u = 37t$. If $t=0$, then $u=10v$ by (12), which according to (3) yields $100a + 10b + c = x = ku^2 = k \cdot (10v)^2 = 100kv^2$, implying $b=c=0$. This contradiction means $t \neq 0$. Moreover sccording to (12) we obtain $37t < 10v \leq 10 \cdot 31 = 310$, yielding $t \leq 8$. Futhermore by (6)-(8) and (12) $(13) \;\; 27a = k(370t^2 - 200tv + 27v^2)$, $(14) \;\; 27b = ktv$, $(15) \;\; 27c = kt(10v - 37t)$. The fact that $k$ is squarefree combined with (14) give us $9 | tv$, which means $3 | v$ (since $1 \leq t \leq 8$). Likewise $9 \mid 370t^2 - 200tv + 27v^2$ by (13), i.e. $(16) \;\; 9 \mid t(t - 2v)$. We know that $3|v$, which means $3|t$ by (16). Hence $t \in \{3,6\}$ (since $1 \leq t \leq 8$). Assume $t=6$. Then $(17) \;\; u = 10v - 222$ by (12), which give us $23 \leq v \leq 25$ since $1 \leq u \leq 31$. Consequently $v=24$ (since $3|v$), which inserted in (14) with $t=6$ result in $3b = 16k$. Hence $16|b$, which is impossible since $1 \leq b \leq 9$.. This contradiction implies $t=3$. Then $(18) \;\; u = 10v - 111$ by (12), which give us $12 \leq v \leq 14$ since $1 \leq u \leq 31$. This combined with $3|v$ and (18) yields $(u,v) = (9,12)$, which according to (6)-(8) means $(19) \;\; (3a,3b,3c) = (2k,4k,3k)$. Hence $3|k$, which according to (19) give us two solutions, namely $(a,b,c) = (2,4,3)$ and $(a,b,c) = (4,8,6)$. Conclusion: The system of equations (1)-(4) has exactly 11 solutions which are $(a,b,c) = (2k,4k,3k)$, where $1 \leq k \leq 2$, and $(a,b,c) = (m,m,m)$, where $1 \leq m \leq 9$.