Question

A B C - C C B = B C A interchanging digits in l
letters

Answer 1

459+495=954

Let’s write ABC algebraically as 100a+10b+c100a+10b+c, and analogously to the other expressions, so we have:

(100a+10b+c)+(100a+10c+b)=(100c+10b+a)(100a+10b+c)+(100a+10c+b)=(100c+10b+a)

Rearranging the terms:

199a+b−89c=0199a+b−89c=0 or 199a+b=89c199a+b=89c

Given that 0≤b≤90≤b≤9, 1≤a≤91≤a≤9, and 1≤c≤91≤c≤9. A first conclusion is that 89c89cmust be close to a multiple of 200. So let’s inspect the first nine multiples of 89: 89,178,267,356,445,534,623,712,80189,178,267,356,445,534,623,712,801.

Only the last of them is close enough to a multiple of 200, so c=9c=9, and the equation is: 199a+b=801199a+b=801. So aa must be 4, and 796+b=801796+b=801, so bb is 55.

So ABC=459=459, ACB=495=495, and CBA=954=954.