For an integer greater than 1, n! denotes the product of all the numbe

Question

For an integer greater than 1, n! denotes the product of all the numbers from 1 to n, inclusive. If x and y are positive integers is x! > y! ?

(1) |x| < a and |y| < b where a < b
(2) When represented on the number line, x is equidistant from 5 and y.

kungfury42 · Answer

Let a = 1000 and b = 2000
Let x = 100 and y = 200
x < y

Let a = 1000 and b = 2000
Let x = 200 and y = 100
y > x

Therefore, we cannot determine a unique relation between x and y. Statement 1 is insufficient. Eliminate options A and D.

y - x = x - 5
x = (y+5)/2

Let y = 1, therefore x = 3. Here, x > y
Let y = 15, therefore x = 10. Here, x > y

Therefore, we cannot determine a unique relation between x and y. Statement 2 is also insufficient. Eliminate option B.

Combining the two statements, we know the cases discussed above in (2) satisfy both the conditions.  Therefore, even the 2 statements together are not sufficient to determine a unique relation between x and y. Eliminate C.

Hence, the question cannot be answered even after using both the statements together, option E.

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JerryAtDreamScore · Answer

Rephrase the question: The factorial function y = x! is an increasing function, so we can rephrase the question as "is x > y ?" Statement 1 Alone: We have x < a and y < b since x and y are both positive. We are allowed to set a and b very high so that the relation between x and y is unrelated to the value of a and b. Then we don't know who is bigger between x and y, insufficient. Statement 2 Alone: This means x is between 5 and y. If y > 5 then we have y > x > 5, but we could also have y = 1 and y < x < 5, so this is still insufficient. Both Statements Combined: We cannot eliminate any cases in statement 2 by combining them with statement 1, so combined it is still insufficient. Answer: E

varshas044 · Answer

Statement 1: |x| < a and |y| < b where a < b It depends on individual value of a and b. We don’t know how much smaller magnitude of x is in comparison to a, and vice versa for relationship between y and b. We will not get any conclusive value of x! > y!? -------- NOT SUFFICIENT Statement 2: When represented on the number line, x is equidistant from 5 and y. X = 4, y = 3 --------- x! > y! X = 6, y = 7 --------- x! < y! X = 3, y = 1 -------- x! > y! I mean you can make as many cases as you want. ---------- NOT SUFFICIENT Statement 1+2: Upon combination we are still stuck with the uncertainty in value as the 2 statements are mutually exclusive. --- NOT SUFFICIENT Answer - E

gmatprepper94111 · Answer

--------------------------------
What do we know right away?
--------------------------------

We just want to know if x, y positive integers tell us x! > y!

Because x and y are positive integers, we may later find it helpful to imagine y = x + k for some integer k. For instance, if k = -1, then x!/y! = x. We'll just stow this away as an observation for later.

-------------------------------
What does statement A tell us?
-------------------------------

There exists some integers a, b such that the following are all true:
--> a < b
--> |x| < a
--> |y| < b

Because x and y are positive integers, |x| = x and |y| = y. We can re-imagine the above statements knowing this.
--> a < b
--> x < a
--> y < b

Note that we can't say much about how x compares to y. Suppose a and b are just huge numbers, and b = a + 1 (We can actually imagine b = a + ε for some arbitrarily small ε!) It's pretty clear we don't know a lot about x and y.

So what does this tell us about x! as compared to y! ? Going back to the point I made a moment ago, not a lot! Let's simplify things a bit and assume a and b are really big numbers. Specifically, let's suppose a = 1,000,000 and b = 1,000,001 so we don't really have to think about them for a moment.

Let's pick our first example. Suppose x = 2 and y = 3.
--> First check against a and b; |2| < 1,000,000 and |3| < 1,000,0001, so we're good.
--> Is x! > y!? No. 2! = 2 and 3! = 3 * 2, so y! > x!.

Let's pick another example. Suppose x = 3 and y = 2. (These numbers are chosen on purpose.
--> A quick check against a and b shows we're okay.
--> And we already know 3! > 2!, so x! > y!.

Statement A is insufficient.

---------------------------------
What does B tell us?
---------------------------------

x is equidistant on the number line from 5 and y. This is equivalent to saying |x -5| = |x - y|

Note that we don't know where x and y lie in relation to 5. It could be true that 5 < x < y or that 5 > x > y. It could also be true that x = y = 5.

Let's try two concrete examples to drive this home.

Suppose x = 5 and y = 5. Then we can observe that x! = y!; so the statement of interest is false.

But what if x = 4 and y = 3? Then we can observe that x! > y!; and the statement of interest is true.

So Statement B is insufficient.

-------------------------------
What about both statements together?
-------------------------------

The fun thing about picking intuitively large numbers for a and b in the first part of the problem is that we can apply the same constraint to our most recent set of counterexamples.

x = 5, y = 5 still works in the sense that 5 < 1,000,000 and 5 < 1,000,0001. And here it's not true that x! > y!.

x = 4, y = 4 still works too. And here it is true that x! > y!.

So both statements together are insufficient.

For an integer greater than 1, n! denotes the product of all the numbe

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