ABC+ACB=CBA solve this equation pls.?

Answer 1

Expanding,

100A + 10B + C + 100A +10C + B = 100C+10B+A

200A + 11B + 11C = 100C + 10B + A

199A + B = 89C

B = 89C - 199A

C=9, A = 4 and B=5

Hence,

ABC+ACB = CBA

459+495 = 954

Answer 2

Abc Cba

Answer 3

ABC = 100A + 10B + C

ACB = 100A + 10C + B

so (100A + 10B + C) + (100A + 10C + B) = 100C + 10B + A

200A + 11B + 11C = 100C + 10B + A

199A + B - 89C = 0

B and C can be any arbitrary integer between 0 and 9 (one digint integer)

and 200A = 89C -B or A = (89C - B)/200 (the range for this function is between 0 and 9)

therefore (89C - B) mod 200 = 0 so that for A we get an integer.

in other words 0 <= (89C - B)/200 <= 9 and (89C - B)/200 must be an integer.

i think there are more than one solution for this equation.

use linear algebra and matrices and basis to find the solutions.

Answer 4

. . ABC
+ ACB
----------
. . CBA

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

A = 4, B = 5 and C = 9. This is the only solution

Answer 5

459+
495=
954

Answer 6

The equation cannot be solved.

First we rearrage our letters to see it more clearly. Since everything is multiplied, each triple letter pair can be rearranged.

so we have

ABC + ABC = ABC

this means

2ABC = ABC

This means there's no solution to the problem except all values being 0.

Answer 7

That's interesting

Answer 8

it is 2abc.

In algebra, a.b.c = a * b * c = c * b* a

so ABC+ACB = 2abc.

so your equation above is not correct.

Answer 9

I too have the same question

Answer 10

a, b, and c all equal zero