If A,B,and C are distant digits in the correctly worked addition problem below,what is the value of A+B+C?
ABB
+9B7 Just to make clear the 9 is under the B on the top row(the one all the way too the right)and the C is
= AA7C is in under but halfway on the B and the 7
A.4 B.9 C.14 D.16
I do not know if this is the best way to solve it but here is I how did it:
We know from the second column that since the second-to-last digit in the sum is odd, the sum from the last rows’ digits must be 10 or greater (but at most 16). Also, we then know that 2B+1 is 7 or 17. Thus, B could be 3 or 8. Now, I also know that A must be 1 because the sum of any two three digit numbers, if four digits long, must start with a 1 (A is not 0 because then we have the sum of a three digit and two digit numbers being a two digit number). Now we will test both cases for B. If B is 3 then the last digit of B+7 is C, so C is 0. Now we check the entire operation and get 133+937=1170, which is not true. Now we check the other case where B is 8. This would make C=5. Testing, 188+987=1175. This is true so A=1, B=8, C=5 and 1+8+5=14 so the answer is C.
Or considering that A=1, we can use algebra to get:
100A+21B+907=1100A+C+70
1000A-21B+C=837
163=21B-C
Since B and C are positive integers less than 1, int(163/21)+1=B=8. Then we can solve for C=-163+168=5.
1+5+8=14