If p and n are positive integers greater than 2, is p^n − p^{n−2} divi

Question

If p and n are positive integers greater than 2, is  p^n − p^{n−2}  divisible by 8?

1) n is odd, and n > 5

2) p is odd, and p > 5

chetan2u · Answer

p^n-p^{n-2}=p^{n-2}(p^2-1)=p^{n-2}(p-1)(p+1)

1) n is odd, and n > 5
We can say n \geq  2a+5, where a is an integer greater than 1.
 p^{n-2}(p-1)(p+1)= p^{2a+5-2}(p-1)(p+1)=p^{2a+3}(p-1)(p+1)=p^{2a}*p^3*(p-1)(p+1) 
 OR simply take the lowest possible value of n, that is 7. 
If p is even, then p^3 is multiple of 8.
If p is odd, then (p-1)(p+1) is product of two consecutive even integers, thus product of 2*4 or 8
Sufficient

2) p is odd, and p > 5
 p^{n-2}(p-1)(p+1) 
(p-1)(p+1) is product of two consecutive even integers, thus product of 2*4 or 8
Sufficient

D

TestPrepUnlimited · Answer

We can factor out  p^{n−2}  and ask if  p^{n -2}*(p^2 - 1) = p^{n-2}*(p - 1)(p + 1)  is divisible by 8.

Statement 1: 
Since n > 5, we have  n - 2 > 3 . Then the power of p is at least 4.

If p is even,  p^{n - 2}  will have at least 4 multiples of 2, then the entire product would be a multiple of 8.

If p is odd, then  (p - 1)*(p + 1) = 	ext{Even * Even} . Note these two evens are consecutive, so we must have a multiple of 4 and a multiple of 2 multiplied together. Then the product must have a multiple of 8.

Sufficient.

Statement 2:

Same as the odd case in statement 1. Sufficient.

Ans: D

If p and n are positive integers greater than 2, is p^n − p^{n−2} divi

Prep Toolkit

Top 5 GMAT Debrief Videos

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards