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Given that p is a positive even integer with a positive units digit, [#permalink]
juangg wrote:
motion2020 wrote:
GeminiHeat wrote:
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.

p=2a is even integer, i.e. when a = 0, 1, 2 ... ... 2a = 0, 2, 4 (excluding a=0, as p is a positive integer)
(2a)^3 - (2a)^2 = 8a^3 - 4a^2 = 4a^2 * (2a - 1) is equivalent to even value (4a^2) having unit's digit as 0 or odd value (2a-1) not possible to have unit's digit as 0. Hence, 4a^2 has unit's digit 0 and we must have a^2 to end with 5 OR a ending with unit's digit 5. Given that p=2a is even, we always have 0 at the end and 0+3 results in 3. Answer is A.


I don't think this is right. The question says that p has positive unit digits, and you said 'always have 0 at the end', which contradicts the former.

If we say p = 6 then we have p^3 = 216 and p^2 = 36, thus the difference in units digits is 0, as we want, and of course 6 is an even positive integer with positive unit digit. Then we have 6 + 3 = 9, i.e. answer D.

for the number p's unit digit yes. I also defined p=2a and a cannot be 0. Please check my post carefully.
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