(x-1)(y-1) = 1?

Statement 2 says x=y. However, x and y could be any number (3,20,-1/2), so we're not sure if it's 1.

A) 16^4 B) (4!)^4 C) 16!/[(4!)^4] D) 16!/4! E) 4^16 I might find it helpful if I work in reverse on some of these more difficult combination problems. So I'm going to try and figure out the description for each answer and see if something matches up to the question at hand. A) 16^4 would apply if w...

It's basically the same thing since there aren't any prime factors between 47 and 50. mapsingh did a nice job of explaining it. One way to look at it is through the divisibility rules. If X is divisible by Y and and Z is NOT divisible by Y, then X+Z will NOT be divisible by Y. Based on mapsingh's an...

I would ignore the diagram. We know that statement 1 doesn't work, because we just have ratios for 3 of the 5 potential areas, but not an actual area measurement for any of the 3 areas. Statement 2 doesn't work, since we're just given the area of A with no relation to D. If we combined them, we know...

The first digit has 4 options (2,4,6,8 and not 0), the second has 5 options (1,3,5,7,9) the third has 3 options (3,5,7 and not 2), the fourth has 7 options (10-3 used before) and the fifth has 6 options (10-4 used before) . The total is 4*5*3*7*6 = [spoiler] 2520 [/spoiler]. As Anurag pointed out, ...

Agreed with the ambiguity. I tried attacking this with the divider method, but my answer was 4.

We'll want to setup an equation where x + xC2 > 12 We can easily eliminate 24 and 12, since we don't need an individual color for each client if we can make pairs. Start with the lowest amount 5 + 5C2 = 5 + 5!/(3!*2!) -> 5 + 5*4/2 -> 5 + 5*2 = 15. Since 15 > 12, we have enough colors with 5 colors t...

We don't know that x = y - z. Statement 1 tells us that x = z - y, once we isolate X. If we multiply all sides by negative 1, we get -x = y - z. So |x| will equal y-z only when x is a negative number, since |x| has to be positive. It's a bit confusing because when we say -x, we're saying that x is a...