The discussion revolves around solving for the three-digit number $\overline{abc}$ given the equation $N=\overline{acb}+\overline{bac}+\overline{bca}+\overline{cab}+\overline{cba}=3194$. Participants explore the relationships between the digits and their permutations, focusing on the mathematical properties of the sums involved.
Participants generally agree on the method of finding $\overline{abc}$ and the calculations involved, particularly regarding the multiples of $222$. However, there is no explicit consensus on the uniqueness of the solution, as the discussion does not resolve whether other values could also satisfy the conditions.
The discussion does not address potential limitations or assumptions regarding the uniqueness of the solution or the completeness of the analysis of other possible values for $\overline{abc}$.
[sp]The sum of all six numbers is $(100 + 10 + 1)(2a+2b+2c) = 222(a+b+c)$, so it has to be a multiple of $222$. Also, it has to lie between $3194$ and $4194$. The only multiples of $222$ in that range are $$\begin{aligned} 3330\ &= 15\times 222 \text{, corresponding to } \overline{abc} = 3330 - 3194 = 136, \\ 3552\ &= 16\times 222 \text{, corresponding to } \overline{abc} = 3552 - 3194 =358, \\ 3774\ &= 17\times 222 \text{, corresponding to } \overline{abc} = 3774 - 3194 =580, \\ 3996\ &= 18\times 222 \text{, corresponding to } \overline{abc} = 3996 - 3194 = 802. \end{aligned}$$ Also, the sum $a+b+c$ must be equal to the corresponding multiple of $222$. The only case where this occurs is when $\boxed{\overline{abc} = 358} = 16\times 222$, with $3+5+8 = 16$.[/sp]Albert said:if $N=\overline{acb}+\overline{bac}+\overline{bca}+\overline{cab}+\overline{cba}=3194$ is the sum of five 3-digit numbers
please find :$\overline{abc}$
(here $\overline{abc}$ is also a 3-digit number)
very nice solution!Opalg said:[sp]The sum of all six numbers is $(100 + 10 + 1)(2a+2b+2c) = 222(a+b+c)$, so it has to be a multiple of $222$. Also, it has to lie between $3194$ and $4194$. The only multiples of $222$ in that range are $$\begin{aligned} 3330\ &= 15\times 222 \text{, corresponding to } \overline{abc} = 3330 - 3194 = 136, \\ 3552\ &= 16\times 222 \text{, corresponding to } \overline{abc} = 3552 - 3194 =358, \\ 3774\ &= 17\times 222 \text{, corresponding to } \overline{abc} = 3774 - 3194 =580, \\ 3996\ &= 18\times 222 \text{, corresponding to } \overline{abc} = 3996 - 3194 = 136. \end{aligned}$$ Also, the sum $a+b+c$ must be equal to the corresponding multiple of $222$. The only case where this occurs is when $\boxed{\overline{abc} = 358} = 16\times 222$, with $3+5+8 = 16$.[/sp]
Opalg said:[sp]The sum of all six numbers is $(100 + 10 + 1)(2a+2b+2c) = 222(a+b+c)$, so it has to be a multiple of $222$. Also, it has to lie between $3194$ and $4194$. The only multiples of $222$ in that range are $$\begin{aligned} 3330\ &= 15\times 222 \text{, corresponding to } \overline{abc} = 3330 - 3194 = 136, \\ 3552\ &= 16\times 222 \text{, corresponding to } \overline{abc} = 3552 - 3194 =358, \\ 3774\ &= 17\times 222 \text{, corresponding to } \overline{abc} = 3774 - 3194 =580, \\ 3996\ &= 18\times 222 \text{, corresponding to } \overline{abc} = 3996 - 3194 = 802. \end{aligned}$$ Also, the sum $a+b+c$ must be equal to the corresponding multiple of $222$. The only case where this occurs is when $\boxed{\overline{abc} = 358} = 16\times 222$, with $3+5+8 = 16$.[/sp]