Question 527848: I have tried many different ways, but i cannot figure out the answer, i have even asked friends and family to help out!
"In the following addition problem, the letters A, B, and C stand for three different digits. What is the sum of the digits that A, B, and C represent?"
ABC
+ ACB
______
CBA
I have tried certain combinations such as 123+132 = 255 , but the "C"'s,"A"'s, and "B"'s do not match up :(
P.S. Thank you in advance
Found 3 solutions by jim_thompson5910, KMST, scott8148:
Answer by jim_thompson5910(35256)   (Show Source):
You can put this solution on YOUR website! This is just something that is done by trial and error.
One thing to point out is that B+C = C+B (since addition is commutative), but notice how C+B = A (in column 3) and B+C = B (in column 2). So this means that a carry is happening. It also implies that B = A+1 (since the largest carry possible is 1)
So because a carry is happening, we know that either B or C is at least 5. In addition, we know that A+A = C. So A must be less than 5 (since C is a single digit number).
These notes help reduce the selection field.
So use trial and error, and stay restricted to the notes above, to find that the answer is
A = 4
B = 5
C = 9
This then tells us that
459
+ 495
______
954
If you need more help, email me at jim_thompson5910@hotmail.com
Also, please consider visiting my website: http://www.freewebs.com/jimthompson5910/home.html and making a donation. Thank you
Jim
Answer by KMST(5328)  (Show Source):
You can put this solution on YOUR website! A is 4 or less because A+A<10 (because the sum has only 3 digits, not 4).
When you add C+B in the ones' place the answer ends in A, but it must be 10+A, because the sum of the tens place (including a 1 possibly carried over) does not en in A, but ends in B.
So B+C=10+A for the ones
and B+C+1=10+B for the tens
That means that B=A+1
and A+A+1(carried over)=C
A=4, gives you B=5, and C=9, and that is one solution.
Is there another one?
If A<4, then B=A+1<4+1=5, and C=A+A+1=not more than 3+3+1=7
B+C<7+5=12=10+A, which would make A = 2 or less, but that would make
B=A+1<4, C=A+A+1=5 or less, and then B+C<9, which does not work.
So there is only one solution, the one listed above.
Answer by scott8148(6628)  (Show Source):
You can put this solution on YOUR website! in the hundreds column, 2 A's add to a C ___ so A is half of C (or less, depending on a carry from the tens column)
in the ones column, C+B gives A (which is less than C) so one must be carried into the tens column
in the tens column, 1+B+C gives B ___ so C=9
with one carried to the hundreds column, 1+A+A gives 9 ___ so A=4
back to the ones column, 9+B gives 4 (with a carry) ___ so B=5
459 + 495 = 954
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