Consider positive integers p and q, such that p q 3. When p! - q!

Question

Consider positive integers  p  and  q , such that  p > q > 3 . When  p! - q!  is divided by positive integer  x , the remainder MUST be zero when  x  is equal to which of the following?

I.  p - q

II.  p - 1

III.  q - 1

(A) I only

(B) II only

(C) III only

(D) I and III only

(E) II and III only

Bunuel · Accepted Answer

GMATNinja Video Solution:

https://www.youtube.com/watch?v=sFIf2ow4mJw

paragw · Answer

If p > q, then we can always take q! common out of p! - q!
Therefore, it'll always have q - 1 as a factor

Rest options are not always true
If you wanna check, you can by taking simple values of p & q like, 5 & 4 or 6 & 4 etc.

Hence,  (C)  III only

SatyP12345 · Answer

The GMAT Ninja video partially confuses me. If you have (x-y)/z and change it to x/z - y/z you can't assume each individual fraction will be an integer. For example if we had (6-1)/5 and changed it to 6/1 - 1/5 it doesn't create two individual integers but clearly the expression 6-1 is divisible by 5.

I understand that if x is divisible by z and y is divisible by z then the expression will be divisible by z. But it seemed they were checking to see if z was a factor of each individually and even if z is not a factor of each individually it can still be a factor of the expression x-y

Appreciate any clarity!

egmat · Answer

Hi SatyP12345,

You've actually spotted a real logic rule, and the good news is the solution never breaks it. Let me line up the two directions so you can see it clearly.

There are two separate statements about divisibility, and only one of them is being used:

-  Valid direction (used for III):  If  x  divides  both a  and  b , then  x  divides  a - b . This is always true, no exceptions.
-  Invalid direction (NOT used):  "If  x  doesn't divide  a  and doesn't divide  b  individually, then  x  doesn't divide  a - b ." This is false - and your  (6 - 1)/5  example is the perfect proof that it's false.

Why statement III is safe

For III,  x = q - 1 . Here's the key point you may have missed:  bothp!  and  q!  are  individually  divisible by  q - 1 .

-  q! = q·(q-1)·(q-2)···1  - it literally contains  (q-1)  as a factor.
-  p! = p·(p-1)···(q+1)·q!  - it contains  q! , so it also contains  (q-1) .

Since  q - 1  divides each of them individually, it must divide their difference. That's the  valid  forward direction - so III is airtight. No leap of faith needed.

Why I and II are rejected

Here the solution does  not  use your feared reverse rule. It just plugs in numbers and computes the actual remainder of  p! - q!  itself:

-  II  ( x = p-1 ): take  q = 4 ,  p = 6 . Then  p! - q! = 720 - 24 = 696 , and  696 ÷ 5  leaves remainder  1 . So II is not guaranteed.

That's a direct check on the difference - exactly the honest way to handle cases where individual divisibility wouldn't tell you anything.

So your instinct is correct, and it doesn't clash with the answer: III is proven  true  by the safe forward rule, while I and II are knocked out by real counterexamples, not by the invalid reverse rule.

Answer: C

Consider positive integers p and q, such that p > q > 3. When p! - q!

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