anceer 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
Official solution from
Veritas Prep.
D
For S1, let's try a few possibilities. 1 + 2 + 3 = 6 is the smallest, and that doesn't work. 1 + 2 + 4 does, so that's a possibility. 1 + 2 + 5 is too big, so that (or anything 1 + 2 + a number bigger than 5, for that matter) is out. The smallest possibility without 1 involved is 2 + 3 + 4 = 9, which is also too big ... so we must have 1, 2, and 4. Hence abc = 1 * 2 * 4 = 8, so S1 is SUFFICIENT.
For S2, algebra doesn't seem to help, so let's use number properties instead.
ab + bc + ca = 14, so ab + bc + ca = even. We can only have an EVEN sum if AT LEAST one of ab, bc, and ca is even. (If they're all odd, we'd have odd + odd + odd, which is NOT even.) If at least one of ab, bc, and ca is even, then at least one of a, b, and c is even. Now let's suppose that ONLY one of them is even: in other words, that a is even, b is odd, and c is odd. Then ab + bc + ca = evenodd + oddodd + odd*even = even + odd + even = odd. So this isn't possible either!
Hence, we know that at least TWO of a, b, and c are even. Now let's suppose that ALL THREE are even. Using the SMALLEST possible even a, b, and c, we'd have 2 * 4 + 2 * 6 + 4 * 6, which is too big ... so they can't all be even!
So we know we have two even and one odd.
1 * 2 + 1 * 4 + 2 * 4 = 14 works, so we could have a * b * c = 1 * 2 * 4
2 * 3 + 2 * 4 + 3 * 4 = 26, which is too big ... so anything other than 1, 2, and 4 is too big. Hence a * b * c = 8, and S2 is ALSO sufficient.
why arent we considering zero here. zero is also an integer and ques doesnt say anything about numbers being greater than zero