Take a quick look at the answer choices. There can't possibly be an infinite number of integer values that would make that expression less than 10. So all we have to do is show that there are more than 4 possibilities, and the answer will have to be D. (And if there are 4 or fewer, it won't take that long to test those.)Mo2men wrote:For how many integer values of x, is |x - 3| + |x + 1| + |x| < 10?
(A) 0
(B) 2
(C) 4
(D) 6
(E) Infinite
If x = 0, we get |0 - 3| + |0 + 1| + |0| = 3 + 1 + 0 = 4. Less than 10. so this works
If x = 1, we get |1 - 3| + |1 + 1| + |1| = 2 + 2 + 1 = 5. Less than 10. so this works
If x = 2, we get |2 - 3| + |2 + 1| + |2| = 1 + 3 + 2 = 6. Less than 10. so this works
If x = 3, we get |3- 3| + |3 + 1| + |3| = 0 + 4 + 3 = 7. Less than 10. so this works
If x = 4, we get |4 - 3| + |4 + 1| + |4| = 1 + 5 + 4 = 10. Not less than 10
We've got four values that work. Let's try a negative
If x = -1, we get |-1 - 3| + |-1 + 1| + |-1| = 4 + 0 + 1 = 5. Less than 10. so this works
So there's more than four possibilities. That means the answer must be D
And if we want to be thorough
If x = -2, we get |-2 - 3| + |-2 + 1| + |-2| = 5 + 1 + 2 = 8. Less than 10. so this works, and our six values are x = -2, -1, 0, 1, 2, or 3
