Given,
p = 2,4,6,8....2n
p
≠ 0 since U.D. is positive p^3-p^2 results in a number whose U.D is 0,
P^2(p-1) should result in a number with U.D. as 0....(1)
Substitute the possible values of p in (1) starting with 2 until you get a number with U.D. as 0
If p=2, 4x1 = 4
If p=4, 16x3 = 48
If p=6, 36x5 = 180 Bingo!
p+3 = 6+3 = 9
u2lover
Given that p is a positive even integer with a positive units digit, if the units digit of p^3 minus the units digit of p^2 is equal to 0, what is the units digit of p + 3?
A. 3
B. 6
C. 7
D. 9
E. It cannot be determined from the given information.