Find the digit number: BAC+BCA = ACB.?

Question

Like to know the procedure to solve the problem.

Answer 1

You have to examine the equations...

First looking at the units column, you know C+A = B or 10+B (if it results in a carry).

Now look at the tens column where you have A+C... this can only result in a C result if you do get a carry, otherwise you should get B like the first column.

So now you know:
A+C = 10 + B <-- units column resulting in a carry
A+C+1 = 10 + C <-- tens column taking the carry and resulting in another carry

Solving the bottom equation you see that the C cancels out from both sides:
A + C + 1 = 10 + C
A + 1 = 10
A = 9

In the hundreds column you have B+B+(1 carry) =A, but you know that A = 9
2B + 1 = 9
2B = 8
B = 4

Finally, plugging these two values (A = 9, B = 4) into the first equation:
A+C = 10 + B
(9) + C = 10 + (4)
9 + C = 14
C = 5

So A = 9, B = 4, C = 5

That makes your equation:

...495
+ 459
--------
...954

Answer 2

now because the permutationa are added we have

A+B+C = 9 or 18 ...1

The reason is take mod 9 we get
(B+A+C)+(B+C+A) = (A+C+B) mod 9

A+B+C = 9 means

C+A = B becuse B < 9 from 1
A+C = C
2B = A this has no sulition

Now let A+B+C = 18 ...1
B < A so C+A = B or 10+B
so C+A = 10+B ...2
so
from 1 and 2 10+2B = 18 or B= 4

if B=4 then as A = 2B or 2B + 1
A = 8 or 9
if A = 8 C= 6
so the numbers are 468+486 means A = 9 which is contradiction
If A = 9 C = 5
495+459 = 954

this is the solution

Answer 3

Well, by logic:
c=b+1
because in the 1 digit A+C = B ( or 10 +B)
in the 10 digit C+A = C (or 10 + C). Assuming B is NOT equal to C (in these problems it always says) c = b+1
and A=9 B=4 C=5 so 495 + 459 = 954
I don't have a procedure. If you want a procedure, try it one by one (using a computer of course).

Answer 4

954