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GRE Prep Club Team Member
Joined: 20 Feb 2017
Posts: 2506
Given Kudos: 1051
GPA: 3.39
Given that p is a positive even integer with a positive units digit,
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28 Jul 2021, 03:08
28 Jul 2021, 03:08
Expert Reply
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Question Stats:
27% (01:21) correct
72% (01:19) wrong based on 11 sessions
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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.
A. 3
B. 6
C. 7
D. 9
E. It cannot be determined from the given information.
Retired Moderator
Joined: 19 Nov 2020
Posts: 326
Given Kudos: 64
GRE 1: Q160 V152
Given that p is a positive even integer with a positive units digit,
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28 Jul 2021, 03:57
28 Jul 2021, 03:57
1
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.
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.
Re: Given that p is a positive even integer with a positive units digit,
[#permalink]
28 Jul 2021, 06:58
28 Jul 2021, 06:58
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.
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.
