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The only positive Even integer which will give same units digit when raised to any power > 0 is 6.. 6^2 will give the same units digit as 6^3..
Therefore the number can be 6,16,26..... Units digit of p+3 will always end with a 9.. D is the right answer.

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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.

given that p^3-p^2 = 0
p^2(p-1)= 0
p is even integer so for above relation only possible when p=6
so p+3 ; 9
IMO D
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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.
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P = Even
P has cyclicity of 1 so. P = 0,1,5,6
Intersection .. P = 6
P+3 = 9...D
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