What is the 3-digit number?

From the left column, we know that $A+B+B+C+C\leq 12$ .

By swapping one of these numbers, the most we can change the sum by is $9$ , as they are single-digit numbers.

Therefore, the right column cannot carry $2$ , as $12+9<23$ , so it must sum to $13$ , and carry $1$ . (It also can't sum to $3$ , otherwise there would be no carry in the left column.)

From this, we can see that the sums of the three individual columns are $11$ , $11$ & $13$ respectively.

As the sum of the left and centre column are equal, we know that $A=B$ , and by comparing the right and centre columns, we can see that $C=B-2$ . Now we can find the values by substituting $B$ in for $A$ and $C$ : $\begin{aligned}B+B+B+(B& -\,2)+(B- 2)=11\\ 5B& -\,4=11\\ B&=3\\ A=B&=3\\ C=B& -\,2=1\end{aligned}\\ \overline{ABC}=\boxed{331}$

The 5 numbers in the sum plus $ABC$ must sum to $222\cdot(A+B+C)$ .

Let $A+B+C=x$ then $222x-1223=ABC$ and make a table:

As we can see the only sum that work is $7$ where $ABC=\boxed{331}$

Just solve the following three Natural Eqs:

(1) 2A+2B+C=13,

(2) 2A+B+2C=11,

(3) A+2B+2C=11.

Get A=3, B=3 and C=1,

Answer=331

Note that the sum of the six numbers $\overline{abc}$ , $\overline{acb}$ , $\overline{bac}$ , $\overline{bca}$ , $\overline{cab}$ and $\overline{cba}$ is $222\left(a+b+c\right)$ . This means that $222\left(a+b+c\right) = 1223 +\overline{abc}.$

Since $100\le \overline{abc} \le 999$ , we have $1323 \le 222\left(a+b+c\right) \le 2222$ , which means that $6 \le a+b+c \le 10$ .

Now note that $222 = 2\times 3\times 37$ , $1223 \cong 2 (\bmod 3)$ and $1223 \cong 2 (\bmod 37)$ . So from the equation $222\left(a+b+c\right) = 1223 +\overline{abc}$ , we have $\overline{abc} \cong -2 (\bmod 3)$ and $\overline{abc} \cong -2 (\bmod 37)$ . This means that $\overline{abc} \cong -2 (\bmod 3\times 37) \cong -2 (\bmod 111) \cong 109 (\bmod 111)$

It is also clear that $\overline{abc}$ is an odd number, so $\overline{abc} = 109 + 2k\times 111$ for some integer $k$ . Now $\overline{abc} \in \{109,331,553,775,997\}$ .

As $6 \le a+b+c \le 10$ , the only possible answer is $\overline{abc} =\boxed{331}$ .

We note that $122A+212B+221{\color{#D61F06}C} = 122\color{#D61F06}3$ . This means that $C$ is odd. Since $122A+212B \ge 334$ $\implies 221C \le 889$ , implying that $C = 1$ or $C=3$ .

When $C=1$ , $122A + 212B = 1223 - 221 = 1002$ , then $61{\color{#D61F06}A}+106B = 50\color{#D61F06}1$ implying that $A$ is odd. It is found that $A=B=3$ satisfy the equation. There is no solution when $C=3$ . Therefore $\overline{ABC} = \boxed{331}$ .

We are given that $\overline{acb}+ \overline{bac}+ \overline{bca}+\overline{cab}+\overline{cba} =1223\implies 122a+221b+212c= 1223$ Rearranging we can obtain the following $2(61a+106b) +221c = 1223$ Since 1223 is an odd number thus $221c$ must be an odd number where $1\leq c < 5$ . If $c=5$ then $1223 -221c < 61a+106b<61$ . Thus we don't considered $c=5$ .

case 1 if $c=3$ then $61a+ 106b=280$ . As 280 is even $61a$ must yield a even number, having possible value $a = 2$ (only). For $a\geq 2$ $\begin{cases} 106b =280-61a =158 < 106b , \ a =2 , \ b > 1\\106b = 280 -61a = 36 < 106, \ a> 2\end{cases}$ Therefore, for $c= 3$ $a,b$ are not integers.

Case 2 if $c=1$ then $61a + 106 b = 501$ implying $a$ to be an odd number and $b\leq 4$ . $106b = (2\times 53 )b = 501 - 61a\implies b = \dfrac{501-61a}{2\cdot 53}\leq 4$ Giving us $61a \leq 77, 61a=183 ,61a= 289,61a = 395$ . Out of these values only true value is $61a=183 \implies a = 3$ and $b=3$ .

Hence $\overline{abc} =331$ .

This solution won’t work if at least 1 digit is 0. But here’s how I did it.

Since all three digits are integers, looking at the rightmost column, we realise that 2A + 2B + C cannot equal to 3, and hence must equal to 13. Following this same train of reasoning, both 2A + B + 2C as well as A + 2B + 2C must equal to 11. By adding them all together, 5 ( A + B + C ) = 35, and hence A + B + C = 7. Substituting this into the previous equations, A + B = 6, A + C = 4 and B + C = 4. This means A = B, therefore A = 3, B = 3 and C = 1. Hence, ABC = 331.

Of course, if any of the digits were 0, it would complicate matters slightly.

So let’s say a digit might be 0. It can only either be B, C or both, since A is the first digit. Looking back at the rightmost column, 2A + 2B + C = 3 or 13. And because both 3 and 13 are odd numbers only B can be 0, as any other alternative will end up with an even number. From there, we realise we can’t get integer solutions for both A and C, so obviously B can’t equal to 0, and none of the digits are 0.

100(A+2B+2C)+10(2A+B+2C)+(2A+2B+C)=1223.

122A+212B+221C=1223

Hence C=1 (C is not 3).

This implies that A=3 and B=3.

Desired answer is 331.

It is first of all worth noting that A , B & C cannot all equal each other (1223 is not a multiple of 5 and if A=B=C then you would effectively be calculating $5*AAA$ ). Also, the main thing to note is that each column of the addition (relating to a power of 10) will produce a sum, related to the answer.

We now move on to analyse the problem in detail, column by column. In the left hand column, since we know that A, B, C > 0 (the first digit of any 3 digit number must be > 0) and can determine that $A+2B+2C < 12$ ( $A+2B+2C=12$ fails when the 3 digit rule is applied to the middle column), therefore A < 8 & B, C < 5 or, in the case that B=C, B, C < 3. In addition, as the left and right hand columns differ by A-C, the right hand column gives $2A+2B+C < 12+A-C$ . Now, the fact that A < 8 tells us that the sum is not 23 but applying the 3 digit rule, it cannot be 3 either, so the sum of the right hand column must be 13. Therefore $2A+2B+C=13$ -> $2(A+B)=13-C$ . As 2(A+B) is even, which implies that 13-C must also be even, C must be odd.

The final column we need to consider is the middle column. Using what we already know, the sum of this column is $2A+B+2C=13-B+C$ and the last digit of the sum is a 1. Recalling that B < 5, this sum is < 21 so it must be 11 (we have already established that it cannot be 1 via the 3 digit rule). This also means that the sum of the left hand column is 11. We apply the same method as before to find that B must be odd. We can now solve via simultaneous equations to find B-C (note that we must choose these two equations such that they eliminate A);

$(1)$ $2A+2B+C=13,$

$(2)$ $2A+B+2C=11,$

$(1)-(2)$ -> $B-C=2$ -> $B=C+2$

But we know that B, C < 5 and that they are both odd. Therefore, $B=3$ and $C=1$ . Substituting into equation $(2)$ yields $2A+5=11$ -> $A=3.$ Thus, the number $ABC$ is $331.$

The original equation is equivalent to $122*A + 212*B + 221*C = 1223$ . We also know $A \geq 1$ . And looking at the unit's place, since $2*A + 2*B + C$ is a number ending in $3$ , $C$ must be odd. It is immediate that $C$ must be $1$ or $3$ .

At this point it is faster to test possibilities than to do more algebra. The solution is quickly found: $A=B=3, C=1$ .

Answer 331. My thoughts:

SUM of right column MUST be 13 (not 23, 33, 43.... because SUM of left column MUST be 12 or 11). So, the only difference between left and right column is one letter C and one letter A, and the difference of its SUM is 2 or 1, therefore, that is the difference between C and A digits. Then, we know that A=B by analysing left and center column (or making the SUM of each column) Hence, as we also know that in order to get an even number in right column SUM, either 3 digits are even, or only one them is, so by replacing 1....2....3, we got the answer 331

First, rewrite the sum as

$100(A + 2B + 2C) + 10(2A + B + 2C) + 2A + 2B + C = 122A + 212B + 221C = 1223$

Next, note that C must be odd, as 1223 is odd and 122 and 212 are even.

For $C=5$ , $5\times 221 = 1105$ but $1105 + 122 > 1223$ so $C<5$

For $C = 3$ , $3\times 221 = 663$ and $122A + 212B = 560$ which means $A+B = 5, 10, 15....$ as both of their last digits are 2. But $122\times 5 = 610 > 560$ so $C \ne 1$

Hence $C = 1$ and $122A + 212B = 1002$ meaning $A + B = 6, 11, 16....$ because both of their last digits are 2. But $11\times 122 = 1342 > 1002$ so $A + B = 6$

The solution is now easily found by either testing values or substituting $A = 6 - B$ into $122A + 212B = 1002$ to give $\overline{ABC} = \boxed{331}$

We have that $100A+10C+C+100B+10A+C+\cdots=122A+212B+221C=1223$ . Testing for a few values of $A$ , $B$ , and $C$ gives the answer of $\boxed{331}$ .

Simple python program does this:

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for a in range(1, 10):
    for b in range(1, 10):
        for c in range(1, 10):
            if 221*c + 212*b + 122*a == 1223:
                print("a =", a, "\nb =", b, "\nc =", c)
                break 

By looking at the 1st column (B+C+A+B+A =13) since 3rd column (A+2B+2C <= 12), by solving the equation (A+2B+2C =11) 1st column,(2A+B+2C = 11) 2nd column,(2A+2B+C = 13)3rd column

Observe that in total we have 5 A's, 5 B's and 5 C's. We make use of the 9 test, so working mod 9: 5(A+B+C) ≡ 1223 ≡ 8. This implies that $A+B+C≡7 \pmod 9$ Or in regular aritmethic: A+B+C is either 7,16 or 25. The first column tells us that A+2B+2C cannot exceed 12, so, even with maximum A=9, 2A+2B+2C cannot exceed 21 and hence A+B+C cannot exceed 10. So now we know: $A+B+C=7$

The last column now tells us that 14-C=10k+3. => C=1 and k=1, which is carried to the middle column.

The middle column now tells us that 15-B=10m+2 =>B=3 and m=1, which is carried to the first column.

The first column now tells us that 15-A=12 =>A=3 $\overline{ABC}=331$

case 1: Let 2A+ 2B + C=3 since ABC is 3 digit number thenA not equal to 0. So only one possible solution A=1, B=0,C=1. (not satisfy the condition) case 2: Let 2A+ 2B + C=13 then 100* ( A+ 2B +2 C) + 10* (2A+ B + 2C) +(2A+ 2B + C) = 100 * 11 + 10 * 11+ 13(=1223) by comparing both the side A+ 2B +2 C =11 --------eqn(1) 2A+ B + 2C =11 --------eqn(2) 2A+ 2B + C= 13 --------eqn(3) eqn(1) - eqn(2) -A+ B = 0 A=B --------eqn(4) 2*eqn(3) - eqn(2) 2A + 3B = 15 ( since A = B ) 2A + 3A = 15 5A=15 A=3 and B=3 from eqn(4) put A and B value in eqn (3) C=1 therefore ABC is 331

Since A is non zero, from the units column we see that either

B = 0 and 2A +C = 3 or 2A + 2B + C = 13

From the tens column we see that the former of these is impossible and we have 2A + 2B + C = 13

Further, 2A + B + 2C + 1 = 2 (impossible) or 2A + B + 2C + 1 = 12

From the hundreds column we have

A + 2B + 2C +1 = 12

So:we have a set of simultaneous equations:

2A + 2B + C = 13

2A + B + 2C = 11

A + 2B + 2C = 11

Solving: ABC = 331

Just looking at the last right column ( B+C+A+B+A=2A+2B+C=3) c has to be an odd number because whatever A and B might be, 2A and 2B becomes even but sum 3 is odd. Thus C has to be odd. C can either be 1 or 3 but not 5,7,9 because the maximum C can be is 3 when A and B are 0. If C=3 then 2A+2B has to be 0 or multiples of 10. The first left column shows that A and B can't be 0 or multiples of 10 ( A+B+B+3+3+x=12). Thus C = 1. If C=1 the 2A+2B has to be 12. Solve this you'll get A=B=3. Thus 331.

We have $122A+212B+221C=1223$ . The fact we are dealing with digits (which means $0 \leq A,B,C \leq 9$ ) makes modular arithmetic very useful here.

$2A+2B-4C \equiv 8$ $\mod{15} \implies A+B \equiv 4+2C$ $\mod{15}$

$5A+4B \equiv 1$ $\mod{13} \implies A \equiv 1-4(A+B)$ $\mod{13}$

(the choices of $15$ and $13$ may look arbitrary, but come from testing modulo $3$ and $5$ - always useful to try small primes - and noting that $13|221$ .) If we know $C$ , we can work out some information about $A+B$ and then about $A$ and $B$ individually. By considering the units column of the sum, we can see that $C$ must be odd; by considering the $100$ s column, we have $A+2B+2C \leq 12$ . Putting these together, $C$ must be $1,3$ or $5$ .

Case $C=1$ :

$A+B \equiv 4+2C=6$ $\mod{15}$ ; since $A,B$ are digits, this in fact means $A+B=6$

$A \equiv 1-4 \cdot 6 \equiv 3$ $\mod{13}$ ; so $\overline{ABC}=331$

Case $C=3$ :

$A+B \equiv 10$ $\mod{15}$ ; so $A+B=10$

$A \equiv 0$ $\mod{13}$ ; so $A=0$ ; but then $B=10$ , contradiction

Case $C=5$ :

$A+B \equiv 14$ $\mod{15}$ ; so $A+B=14$

$A \equiv 10$ $\mod{13}$ ; no solution with $0\leq A \leq 9$ ; contradiction

So the answer is $\boxed{331}$

2(A+B) + C = 13 2(A+C) + B = 11 2(B+C) + A = 11

Solving this system gives: A=3 B=3 C=1

$\begin{aligned} a,b,c &\in \mathbb{Z^+} \\ 122a+212b+221c&=1223 \\ 122(a-1)+212(b-1)+221(c-1)&=668 \end{aligned}$
So $c-1$ must be even.
$c-1 = 2 \implies 122(a-1)+212(b-1) = 226,$ which isn't possible, so $c - 1 = 0$ .
$122(a-1)+212(b-1) = 668$ which has a solution $(a-1, b-1)=(2,2)$ .
Therefore, $\overline{abc}=331$ .