sud21 wrote:a, b, and c are three distinct positive integers. What is the product abc?
1. a + b + c = 7
2. ab + bc + ca = 14

we are told that a,b,c > 0 and these 3 are distinct

statement 1
There is only 1 way to have 3 positive numbers to get a sum of 7 : -> 1 and 2 and 4
it doesn't mater which is a or b or c since we want abc
abc = 1 x 2 x 4 = 8 ; statement is sufficient

statement 2
ab + bc + ca = 14
a(b+c) + bc
we can substitute numbers in the equation
of all the choices 1,2,4 satisfies the equation

ans = d

Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem.
Remember equal number of variables and independent equations ensures a solution.

a, b, and c are three distinct positive integers. What is the product abc?
1. a + b + c = 7
2. ab + bc + ca = 14

in the original condition there are 3 variables (a,b,c), and thus we need 3 more equations to match the number of variables and equations. Since there is only 1 each in 1) and 2), there is high probability that E is the answer. Using both 1) & 2) together, we need to focus on the distinct positive integers of the original condition. 1) is satisfied by (a,b,c)=(1,2,4),(1,4,2),(2,1,4),(2,4,1),(4,1,2),(4,2,1), and all of them match the condition abc=8 therefore the answer is unique and the condition is sufficient. Therefore the answer is D. This is also related to common mistake type 4(B).

for 95% of DS questions where 1) = 2), the answer is usually D.


Normally for cases where we need 3 more equations, such as original conditions with 3 variables, or 4 variables and 1 equation, or 5 variables and 2 equations, we have 1 equation each in both 1) and 2). Therefore E has a high chance of being the answer (especially about 90% of 2by2 questions where there are more than 3 variables), which is why we attempt to solve the question using 1) and 2) together. Here, there is 80% chance that E is the answer, while C has 15% chance. These two are the key questions. In case of common mistake type 3,4, the answer may be from A, B or D but there is only 5% chance. Since E is most likely to be the answer according to DS definition, we solve the question assuming E would be our answer hence using ) and 2) together. (It saves us time). Obviously there may be cases where the answer is A, B, C or D.

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S1::

Since a, b, and c are distinct positive integers, the minimum possible sum is 1 + 2 + 3, or 6. We have 7, so we must increase one of these numbers by 1. This can only be 1 + 2 + 4 (anything else could give us a duplicate, e.g. 1 + 3 + 3 or 2 + 2 + 3), so this is SUFFICIENT.

S2::

ab + bc + ca = 14

If a, b, and c are all greater than 1, then we have at least 2*3 + 3*4 + 4*2, which is too big, so one of the numbers must = 1. Let's suppose that it's a. This gives us

b + bc + c = 14, or

(b + 1)(c + 1) - 1 = 14, or

(b + 1)(c + 1) = 15

This has two integer solutions: 1 * 15, and 3 * 5.

Since neither b nor c can be 14, as that alone would give us the whole sum. Our only integer solution is therefore b = 2, c = 4, or vice versa, and this is again sufficient.