MHB Solve for $\overline{abc}:$ $N=3194$

[Albert1]

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)

 

[Opalg]

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)

[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]

 

[Albert1]

[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]

very nice solution!

 

[kaliprasad]

[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]

almost same method solved at

Spoiler

Fun with maths: If abc is a three digit number and a, b, c are distinct digits, what is abc if acb+bca+bac+cab+cba=3194