If A, B and C are digits and AB 0, what is the value of B?

AB#0 means \(AB\neq{0}\), the product of A and B does not equal to zero. This way we are told that neither A nor B equals to zero.

As for the solution it's given in above post. Please ask if anything remains unclear.

I think the answer is D.

Statement 1: this statement has already been established as sufficient in the previous posts.

Statement2: A=1

AB
+BA
-------
AAC


In the above addition, observe the units place: (digit) B + (digit) A = (digit) C
Now observe the tens place: (digit) A + (digit) B = (digit) A? -----> this means that a digit has carried over from the addition of the units digits. Now, because a digit has carried over, this implies that the digit A+B is a two digit number----> Now, the only way A+B or (1+B) will be a two digit number is if B = 9.

therefore statement B is sufficient as well.

You cannot use information given in the first statement (AB + BA = AAC) for the second.

Check complete solution here: if-a-b-and-c-are-digits-and-ab-0-what-is-the-value-of-b-130419.html#p1071580

Hope it helps.

If A, B and C are digits and AB#0, what is the value of B?

(1)
AB
+BA
________
AAC

Important takeaway here is that for the tens digits: A+B results in AA. This means that:

1. A+B is at least 10

Therefore the B+A in the units place is at least 10 and we have to carry a 1 over to the tens digits (cannot be 2 since A+B is at most 9+9=18)

2. A+B in the tens places has a units digit of A

This means that A+B+1 (carried over from B+A in the units places) has a units digit of A. To get a units digit of A, B+1 must be 10.

Therefore B=9

SUFFICIENT

(2) A=1

We know A is 1 but we have no idea what B is

INSUFFICIENT

Answer is A

Algebraic really works fast for this kind of question:

A=x, B=y

it means

y+y=10x+x, 2y=11x

since 1<=x<=9, 0<=y<=9

so this possibility can't rise. there must be one from z+x=c

so 2y+1=11x

so x=1, y=5

sufficient

1. AB +BA = AAC

10A+B+10B+A=100A+10A+C
11(A+B)-100A+10A=C
11(A+B)-110A=C
11(A+B-10A)=C
11(B-9A)=C

Since C is a digit, it can not be 11, so the only one option is when (B-9A)=0, so C=0. In this case the only way is B=9 and A=1. Sufficient.

2. A=1

Not Sufficient

Answer A.

The second statement is sort of a hint for the first, as to what numbers you should try out, because it's outright clear that second itself is insufficient.

so if A is 1, C must give some number that gives off a carry to the next ( that's the only reason one can explain how B + A = C and A+ B = A, because if we didn't have an extra carry, both should've been C or both A) so C is a number giving some carry to the next. , the next, part, (taking some hints from second statement is easier now, so if A=1 then c = A + B = 1+9; c=0 . 1 carry to (A + B) = 1+ 9 + carry 1 = 11 ( 1 carry goes to the next and final phase) the whole sum comes out as = 110.

AB
BA
------
AAC = 110

"(A can not be more than 1, as the sum of 2 two-digit numbers is always less than 200)"
"(B can not be less than 8 as 18+81=99, not a three digit number)"
Hi, i'm confused about this, why is it that a cannot be more than 1 or b cannot be less than 8, please explain in layman terms if possible.

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