If the digits A, B, and C are added, the sum is the two digit number AB as shown below. What is the value of C?
A + B + C = AB
1 + 2 + 9 = 12
Hint: What’s the larger possible sum of three digits? How does that limit possible values of A, B and C?
Answer to the problem:
[SPOILER] A must be either 1 or 2 (because the sum of three distinct digits is at most 24 (I’m assuming distinct digits are intended, ie. A doesn’t equal B or C, and B doesn’t equal C).
But if A is 1 or 2 then the most the sum could really be is 19 (2+9+8), which leads to a contradiction. So A is 1. After that it’s easy - the only way to have A+B+C to equal AB with A=1 is if C=9. Then B can be 2, 3, 4, 5, 6, 7 or 8:
1+2+9=12, 1+3+9=13, etc.
[/SPOILER]
Actually it works for any value of B.
Hint: if A & B & C were all 9, the total would only be 27. So A has got to be a 1 or a 2. Mathematically speaking, A could also be zero, but that’d be a pretty crappy trick to stick in a puzzle, so 1 or 2 it is.
ETA: No other answers were there when I started; must be busy tonight.
The trickery here comes from the fact that they wrote the equation incorrectly. It’s not that A + B + C = AB, but rather that A + B + C = 10A + B. Cancel, and you have that C = 9A. The only way that works for single digits is if A = 1 and C = 9. After that, you’ll find that there are no new restrictions on B.
ETA: You could also have both A and C be 0, but that’s somewhat at odds with the notion that 10A + B is a two-digit number.
Since one specific answer has been given here’s how I approached it:
a+b+c=ab
a+b+c=10a+b
a+c=10a
c=9a
So the original equation becomes:
a+b+9a = 10a+b
10a+b = 10a+b (Which is true for any values of a and b).
There’s an infinite number of solutions. Pick a value for “a”. That gives you “c”, and then you can choose anything you like for “b” - it makes no difference.
ETA - unless you play by the rules as ultrafilter noted and stick with single digits for a, b and c.
Thanks everybody!