If a, b, c are integers such that b>a, is b+c>a?

(1) c>a

(2) abc>c





OA is C

I honestly hate experimenting with numbers because that will only make me feel even more confused. How can we manipulate with the variables algebraically? Here is what I have done so far:


b+c>a
- b>a
c>0, so what the question is really asking is whether C is positive



(1) we can't tell whether both the variables are even positive

(2) worst case scenario, only one of the variables could be positive while the other 2 variables being negative, but we don't know which variable is suppose to be positive

(1&2) Since b>a and c>a, because a is smaller than both b and c, then a must be negative. What we need to figure out is whether c or b is positive. But I don't know where to go from there....

no, you can't subtract inequalities that are facing the same way. no dice.
it's perfectly possible for both b and b + c to be greater than a, even though c is negative. for instance, if b = 10 and a = 5, then c can be any negative number between -5 and 0 and b + c will still be greater than a.

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i think a certain amount of 'experimenting with numbers' on these sorts of problems is inevitable. in fact, believe it or not, that's how real mathematicians solve these sorts of problems: they just go through cases one at a time, meticulously trying out all the different possibilities of positive vs. negative. they don't usually plug in numbers, but it's the same idea.

statement (1)
in this case, you know that each of b and c is greater than a, but that's all you know.
if b and c are positive, then this is good enough, because then b + c will be greater than either b or c (both of which are already greater than a). so that's a Yes.
with negative values, though, you can get a No. if b = -2, c = -3, and a = -4, then b > a and c > a, but b + c < a. that's a No.
insufficient.

statement (2)
this statement rules out c = 0 (because 0 > 0 is impossible), so we should consider two separate cases: c is positive and c is negative.
if c is positive, then divide by it and don't flip the sign, giving ab > 1 (and c > 0). this means, among other things, that a and b have the same sign, and c > 0.
if c is negative, then divide by it and flip the sign, giving ab < 1 (and c < 0). note that ab is 0 or less, because a and b are integers (so no values between 0 and 1). this means, among other things, that a and b have opposite signs or one of them is zero, and c < 0.
yuck.
now pick numbers:
a = 1, b = 2, c = 3: Yes
a = -1, b = 0, c = -2: No
insufficient

(together)
let's revisit the 2 cases above for statement (2).
if c > 0 and ab > 1, then all three of a, b, and c are positive. this means Yes to the question prompt, for reasons detailed above.
if c < 0 and ab < 1, then there are 2 sub-cases:
--- one of a and b is zero: in this case it must be b, because a is less than c (and therefore negative). but if b = 0, then the question prompt becomes 'is c > a?' which is a Yes (because that's statement 1).
--- a and b have opposite signs: in this case b is positive and a is negative, because b must be bigger than a. but we already know c > a, so adding 'b' (which is positive) guarantees that b + c > a. so that's another Yes.
Yes + Yes + Yes = sufficient

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this problem was brutal. there is probably some really elegant solution that's eluding me at the moment.

(2) should read abc > 0

Kevin Armstrong
Madrid, Spain

oh wow.
that certainly makes things much easier.

solution to revised problem:

remember that if c is positive, then the statement is guaranteed to be true (because b is already greater than a, so adding something positive will keep it that way).

(1) is still insufficient for the same reasons i posted above.

(2) means one of the following: (a) all three are positive, (b) two are negative and one is positive.
if all 3 are positive, then a fortiori c is positive, so b + c must be greater than a.
but
if b is positive, and a and c are negative, then it's possible that b + c is not greater than a. if you don't like plugging actual numbers, then consider the idea that c could be a REALLY BIG NEGATIVE that cancels out the positive-ness of b. for instance, if a = -1, b = 2, and c = -100, then b > a, but b + c <<<< a.
insufficient.

(together)
in this case, a is the smallest of the 3 numbers.
3 cases:
* all positive: this is a yes, as established before
* a < b < 0 < c: this is also a yes, because c is positive
* a < c < 0 < b:
rearrange the inequality to "is b > a - c ?"
in this case, notice that b is positive and a - c is negative, so this is still a yes.
always yes
sufficient

ans = c

if anyone has a more direct or more elegant solution, feel free to post it.