Arc from Car

The sum is as follows:

$\begin{aligned} 111C + 11A + R & = 100A + 10R + C \\ 110C & = 89A+9R \\ & = 80A + 9A+9R \\ & = 80A + 9({\color{#3D99F6}A+R}) & \small \color{#3D99F6} \text{Last column: } R+A + C = 10+C \implies A+R =10 \\ & = 80A + 90 \\ \implies 11C & = 8A+9 \\ 11{\color{#3D99F6}C} - 9 & = 8A & \small \color{#3D99F6} \text{Since } 8A \text{ is even and }> 0, C \text{ must be odd and }> 1 \\ {\color{#3D99F6}33} - 9 & = 8A & \small \color{#3D99F6} \text{No integer solution other than }C=3 \\ \implies A & = 3 \\ \implies R & = 10-A = 7 \end{aligned}$

$\implies \overline{ARC} = \boxed{373}$ .

From the ones column, we get $A + R = 10$ . Now 1 is the carrying digit of ones row.

Thus, we get from the tens column $A + C + 1 = R$ or $A + C + 1 = 10 + R$ .

From the hundreds column we get:

If $A + C + 1 = R$ , $A=C$ . As $A + R = 10$ we get $A+C+1 = 10 - A$ and $2A + C + 1 = 10$

If $A + C + 1 = 10 + R$ , $A = C+1$ . As $A + R = 10$ we get $A+C+1 = 20 - A$ and $2A + C + 1 = 20$

This means, either of the following statements holds true:

$A = C$ and $3A + 1 = 10$

OR

$A = C+1$ and $3A = 20$

As 3 doesn't divide 20, there's no solution to the second case. In the first case, we get $A=3$ and thus $C=3$

From $A+R=10$ we get $R+3 = 10$ and $R=7$ .

By checking, we get $337+33+3 = 373$ . As $C=3, A=3, R=7$ is the only possible solution, the answer is $\boxed{373}$ .

This took me awhile to solve. I did not like the fact that they used two different variables for the same number.

I initially thought $C+1=A$ and $A+C+1=10+R$ and that $R+A=10$

These were my three equations I used and set up a system of equations

The answers I got, after trying multiple methods of solving, were not integers. So I reexamined the equations

I went back and realized that $R+A=10$ has to be true.

There is no other way of getting $C$ in the ones place with $C$ being one of the digits added to it. As long as the other digits are non-zero

I next went back to reexamine $A+C+1=10+R$

$A+C+1$ has to be true because since the ones place give us something higher than $10$ we have to carry the $1$ over to the tens place

However getting $10+R$ as a result does not have to be the case if $C=A$

This would mean that $A+C+1=R$ and that nothing carries over to the hundreds place

So now that I modified the equations to: $C=A$ ,

$A+C+1=R$ , and

$R+A=10$
the system of equations could be modified to

$A+A+1=R$

and

$R+A=10$

if $2A+1=R$ and $A+R=10$ then $A + (2A+1)=10$

$3A+1=10$

$3A=9$

$A=3$ , $R=7$ , $C=3$

From the ones column $R + A + C = \text{(10s digit)} + C .$ Subtracting $C$ on both sides, $R + A = \text{(10s digit)} ,$ which implies (since only two digits are being added) that $R + A = 10 .$

This means the tens column has a carry of 1.

Suppose $C \neq A ,$ so we know there is a carry digit of 1 from the tens column.

Combining this information and subtracting $C$ from both sides of the addition leaves us with:

$\begin{array} {cccc} \large & & 1 & 1 & \\ \large & & \color{#D61F06}C & \color{#3D99F6}A & \color{#20A900}R \\ \large + & & & \color{#D61F06}C & \color{#3D99F6}A \\ \hline \large & & \color{#3D99F6}A & \color{#20A900}R & \color{#69047E}0 \end{array}$

This gives us the equations $1 + C = A ,$ $1 + A + C = 10 + R ,$ and $R + A = 10 .$ Doing some substitution leads to the equation $2A = 10 + 10 - A$ which reduces to $3A = 20 .$ Since $A$ is an integer, this is impossible, and our supposition that $C \neq A$ is incorrect.

So $C = A :$

$\begin{array} {cccc} \large & & & 1 & \\ \large & & \color{#D61F06}C & \color{#3D99F6}C & \color{#20A900}R \\ \large + & & & \color{#D61F06}C & \color{#3D99F6}C \\ \hline \large & & \color{#3D99F6}C & \color{#20A900}R & \color{#69047E}0 \end{array}$

This implies $1 + C + C = R$ and $R + C = 10 .$ Substitution implies $1 + C + C + C = 10 ,$ or $3C = 9 .$ So $C = 3,$ $A=3,$ and (using $R + C = 10$ again) $R = 7 .$

$0<A,C, R <10$

From the first column :

• We have $R + A + C < 10$ or $R + A + C > 10$

  • If $R + A + C < 10$ then : $\left\{\begin{matrix} R + A + C = C \\ A + C = R \\ C = A \end{matrix}\right.$ $\implies 3A = 0 \implies A = 0$

So $R + A + C > 10 \implies R + A + C = 10q + C \Leftrightarrow \boxed{R + A = 10q \ , \ q = 1 \ or \ q = 2}$

• Now $A + C + q > 10$ or $A + C +q < 10$

  • If $A + C + q > 10$ Then we would have : $A +C + q = 10q' + R, q' = 1$ since $A < 10$ and $C <10$ So $A+C + q =10 + R$

$\left\{\begin{matrix} R + A = 10q \\ A + C + q =10 + R \\ C + 1= A \end{matrix}\right.\implies 2C + q - 9 + A = 10q \implies 3A = 9q - 11 , \left\{\begin{matrix} 3A = -2 \ if \ q = 1 \\ 3A = 7 \ if \ q = 2 \end{matrix}\right.$ Impossible

So $\boxed{A + C +q < 10}$

And we wloud have :

• $\left\{\begin{matrix} R + A = 10q \\ A + C + q =R \\ C = A \end{matrix}\right.$ $\implies R = 2A +q \implies A = 9q$

$\left\{\begin{matrix} 3A = 9 \ if \ q = 1 \implies \boxed{A = 3} \\ 3A = 18 \ if \ q = 2 \implies A = 6 \implies C = 6, R = 20 - 6 = 14 \ :\ impossible \end{matrix}\right.$

So : $\left\{\begin{matrix} R + A = 10 \\ A + C + 1 =R \\ C = A = 3 \end{matrix}\right. \implies \boxed{ A = 3, C = 3, R = 7, ARC = 373 }$

From the above information, we get three important equations: First: 110C = 89 A + 9R (From 111C + 11A + R = 100A + 10R + C) Second: R+A = 10 or R = 10 - A Thrid: A+C+1=R

From Second and Third, we get: A+C+1 = 10 - A and therefore, C= 9 - 2A.

Now, let's substitute C and R with As and we get:

 110 (9 - 2A) = 89A + 9 (10 - A)

Then we get: 990 - 220A = 89A + 90 - 9A Then we get: 900 = 300A Finally: A = 3, therefore R= 7, therefore, C= 3

I created a python code which checked every possible solution after i couldn't figure it out after 5 minutes of trying. (Yes i gave up quick xD still didnt look at the answer after i got it.) https://pastebin.com/76sRZWV0

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for i in range(1,10):
    for j in range(1,10):
        for k in range(1,10):
            car = 100*i+10*j+k
            ca = 10*i+j
            c = i
            arc = 100*j+10*k+i
            if car+ca+c==arc:
                print(arc)