VeritasPrepKarishma wrote:
petercao wrote:
AB + BA = AAC
In the correctly worked addition problem shown, where
the sum of the two-digit positive integers AB and BA is
the three-digit integer AAC, and A, B, and C are different digits, what is the units digit of the integer AAC?
A. 9
B. 6
C. 3
D. 2
E. 0
Let's see what the addition should look like:
A B
+B A
--------
A A C
Look at the big picture first. Two 2 digit numbers get added to give you a 3 digit number, AAC. A has to be 1. It cannot be 0 because then AAC is not a three digit number and it cannot be anything more than 1 because 2 two digit numbers cannot make 200 or more (To make 200, at least 1 number must be a 3 digit number. The maximum you can make out of the sum of two digit numbers is 99+99 = 198)
So A = 1.
A B
+B A
--------
A A C
Next we see that B+A gives us something ending in C but A + B gives us something ending in A. This means that B+A must have given us a carryover 1.
(1)1 B
+ B 1
-------
1 1 C
To get a carryover 1, B has to be 9 so that sum = 9 + 1 = 10. Therefore, C must be 0.
karishma, thanks for your answer. Just a question about the thought process. I started off thinking AA must be 11,22,33,44,55, etc. then realizing the sum can only be in the 100s as you did yourself.
then I end up with :
1B
+ B1
11C
From here you were talking about how 1+B should = B +1 unless there is a carryover. But shouldn't we have taken a similar, more heuristic approach like we did earlier with the < 200 to think that 1B and B1 cannot be anything less than 9 because 18 + 81 (where b is 8) is already less than 100. then b must be larger and thus 9.
I'm just curious if when you do problems like this, you're actually thinking of the carry, etc. because that seems very difficult to replicate when you are really taking a test.