TO find: p+3

p = +ve _ _ ==> +ve unit digit means that its not "0".. So possible 2,4,6,8

p^3 - p^2 = 0

p^2 ( p - 1) = 0
Since, unit digit of "p" is not 0.. that means the resultant figure has unit digit as "0"

If we quickly scan answer choices:

{A} 3 ==> p = 0 NOT POSSIBLE
{B} 6 ==> p = 3 ==> Not Matching with 0 unit digit
{C} 7 ==> p = 4 ==> Not Matching with 0 unit digit
{D} 9 ==> p = 6 ==> Possible

[spoiler]{D}[/spoiler]

sparkles3144 wrote:Given that p is a positive even integer with a positive units digit, if the units digit of p3 minus the units digit of p2 is equal to 0, what is the units digit of p + 3?

A. 3
B. 6
C. 7
D. 9
E. Cannot be determined

Answer D

Isn't 0 an even integer?

I thought 6^3-6^2 units digits is equal is 0
and 0^3-0^2 = 0
So I chose E.
Why am I wrong?

Please explain. Thank you!

"0" is an Even Integer but NOT Positive.

Question says "positive Unit Digit" that means 1 to 9.. Since, number is Even.. so 2,4,6,8

The key here (as Rahul says) is that p is EVEN (so its units digit is 0, 2, 4, 6, or 8) and has a POSITIVE units digit (so its unit digit is 2, 4, 6, or 8 - 0 is out!)

Now we can test a few values of p in the equation (units digit of p)³ - (units digit of p)² = 0. (The expression (...4) I use below means "an integer that ends in 4".)

2³ - 2² = 4 (nope!)
4³ - 4² = (...4) - (...6) (nope!)
6³ - 6² = (...6) - (...6) (success!)
8³ - 8² = (...2) - (...4) (nope!)

So p must have a units digit of 6, and p + 3 must end in 9.