Question

if each of the four letters in the sum below represents a different digit, which of the following cannot be a value of A?

AB
+ BA
------
CDC

1. 6
2. 5
3. 4
4. 3
5. 2



How do you get the answer to this easily? Can anyone explain it thoroughly?

Answer 1

Try setting A to each of those numbers, where A is different from B,C, and D.

You should find that only one A value is possible.

Answer 2

thanks but i tried that and i wasn't very successful. My book says that voice 5 is correct but i don't understand

Answer 3

You can also solve this with a system of equations I believe.

Answer 4

Hmm. This problem is kinda tricky. Hold on while I think about this a minute.

Answer 5

Well I misread the question for one thing XD.

ONE of those options is what A CANNOT be.

So from the right-most collumn: B + A = C + k

The next collumn: [A + B + k) ]%10 = D

where k is what's carried from the first collumn and %10 is the remainder when divided by 10.

Answer 6

thank you so much

Answer 7

Did you find an answer? I'm still trying to work through it.

Answer 8

i think the easiest way is to plug in for the numbers. I started with the last choice and plugged 2 for A and 8 for B and see why 2 is the answer

Answer 9

but my answer changed when i plugged in 4..that's why I'm confused

Answer 10

Using your method:

2B
+B2
----
CDC

I don't see why any A value is impossible...

Answer 11

Bah, I refreshed the page and my long post was lost. Hold on one more second... XD

Answer 12

I think C must be 1 because:

1) you're using a base-10 number system(when you add two things that make ten you carry to the next collumn)

2) adding two 2-digit numbers where A and B are different, the largest carry you can have in the 2nd collum is 1.

Proof: Let A = 9, and B = 8 (Or vise versa) (8 and 9 are chosen because they are the biggest numbers in a base-10 number system)

89 + 98 = 187

Answer 13

So in my equations above k can only equal 0 or 1 and C = 1.

This still leaves A, B and D. with only two equations...

Answer 14

got it!

Answer 15

How? I'm perplexed by this question!

Answer 16

i plugged in random numbers until i got my answer. Sadly, this long problems is a PSAT sample question! It would've been miserable if i spent all my time on this problem.

Answer 17

http://openstudy.com/study#/updates/4e2fa36d0b8ba7b2da403dbf

Looks like this has been asked before XD

Answer 18

Looks like there is no easy way to answer this question without trying all of the possibilities.

But don't be discouraged, I'm in third year university XD.

Answer 19

Thanks a lot..i guess all of the choices give you 121 except the last one