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Given that p is a positive even integer with a positive

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Given that p is a positive even integer with a positive [#permalink] New post 08 Aug 2006, 12:59
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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.
[Reveal] Spoiler: OA

Last edited by Bunuel on 29 Sep 2013, 08:47, edited 1 time in total.
Renamed the topic, edited the question and added the OA.
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 [#permalink] New post 08 Aug 2006, 13:15
A

Let x is the unit digit of p. Then

Unit digit of p^3 = unit digit of x^3
Unit digit of p^2 = unit digit of x^2

Their difference is 0 (As given). This is possible only if unit digit is 1 or 0. But since p is an even integer. So unit digit of p =0
hence unit digit of p+3 = 3
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 [#permalink] New post 08 Aug 2006, 13:15
Should this be E?

(Unit digit of p^3) - (Unit digit of p^2) = 0 if the units digit is 0, 1, 5, 6.

The answer in these cases would be 3, 4, 8, 9.

Even if 0 is neither posive or negative, we still have 3 options left.

However, only 9 is mentioned from among these choices, so could be D as well... any suggestions? :roll:
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 [#permalink] New post 08 Aug 2006, 13:24
E It cannot be determined from the given information.

As p is even,
p = 10x+y

y = {0,2,4,6,8}

Unit's digit of p^3 = {0, 4, 6}
Unit's digit of p^2 = {0,2,4, 6, 8}

As unit's digit p^3 -unit's digit p^2 =0

Common units digit between p^3 and p^2 = {0, 4,6}

We can get {0,4,6} in p^2 or p^3 from {0,2,4,6,8}

So cannot determine.
Answer: E
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 [#permalink] New post 08 Aug 2006, 13:30
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ps_dahiya wrote:
A

Let x is the unit digit of p. Then

Unit digit of p^3 = unit digit of x^3
Unit digit of p^2 = unit digit of x^2

Their difference is 0 (As given). This is possible only if unit digit is 1 or 0. But since p is an even integer. So unit digit of p =0
hence unit digit of p+3 = 3

What the heck I was thinking :wall

Answer should be D because last digit of p must be +ve and even.
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Re: PS: Digits again [#permalink] New post 08 Aug 2006, 13:42
u2lover 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.

please explain


unit digits could be 0, 1, 5 and 6 but only 6 satisfies the conditions given in the question. so D make sense..
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 [#permalink] New post 08 Aug 2006, 14:46
From the condition,

p is even and not divisible by 10 (because units digits is positive)

so p=2k, and k can not be multiple of 5 or 10

p^3-p^2 = p^2(p-1) = (2k)^2*(2k-1)

Since we know that units digit of p^3-p^2 is 0 and k con not be 5 or 10,
the only way to get 0 at the end of the number is if 2k-1=5

=> 2k=6 => p=6 => p+3=9

(D)
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 [#permalink] New post 09 Aug 2006, 16:03
Hey U2...
Can you tell me where I missed this question?

My solution is below... goes without saying got it wrong...

haas_mba07 wrote:
E It cannot be determined from the given information.

As p is even,
p = 10x+y

y = {0,2,4,6,8}

Unit's digit of p^3 = {0, 4, 6}
Unit's digit of p^2 = {0,2,4, 6, 8}

As unit's digit p^3 -unit's digit p^2 =0

Common units digit between p^3 and p^2 = {0, 4,6}

We can get {0,4,6} in p^2 or p^3 from {0,2,4,6,8}

So cannot determine.
Answer: E
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Re: PS: Digits again [#permalink] New post 09 Aug 2006, 16:13
u2lover 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.

please explain


If units digit = 6 then units(216)-units(36) = 0
p+3 = 9

Hence D

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 [#permalink] New post 09 Aug 2006, 16:20
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haas_mba07 wrote:
Hey U2...
Can you tell me where I missed this question?

My solution is below... goes without saying got it wrong...

haas_mba07 wrote:
E It cannot be determined from the given information.

As p is even,
p = 10x+y

y = {0,2,4,6,8}

Unit's digit of p^3 = {0, 4, 6}
Unit's digit of p^2 = {0,2,4, 6, 8}

As unit's digit p^3 -unit's digit p^2 =0

Common units digit between p^3 and p^2 = {0, 4,6}

We can get {0,4,6} in p^2 or p^3 from {0,2,4,6,8}

So cannot determine.
Answer: E


Since Q says unit digit has to be +ve 0 is out
if 4 is units digit and say # is 4--> 4^3 = 64 & 4^2=16 . Subrtacting units digit does not result in 0.

Try the rest. Only when 6 is in the units digit will 6^3= 216 , 6^2=36. Subtracting units digit will result in zero.

Hence reqd answer = 6+3 = 9 Choice D

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 [#permalink] New post 09 Aug 2006, 16:37
yes... zero is neither positive nor negative so you must leave it alone

I just picked p=2, 4, 6 and got p^3-p^2= 4, 8, 0

p=6 and p+3=9
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 [#permalink] New post 11 Aug 2006, 02:50
P is a even number with a positive number in the unit digit
Hence unit digit can be 2,4,6 or 8

2^2 = 4 2^3 = 8
4^2 = 6 4^3 = 4
6^2 = 6 6^3 = 6
8^2 = 4 8^3 = 2

Hence only 6 in the unit digit satisfies the condition given.

hence unit digit of p+3 = 9
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Re: Given that p is a positive even integer with a positive [#permalink] New post 28 Sep 2013, 11:31
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Re: Given that p is a positive even integer with a positive [#permalink] New post 29 Sep 2013, 08:57
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u2lover 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 is a positive even integer with a positive units digit --> the units digit of p can be 2, 4, 6, or 8 --> only options C and E are left.

In order the units digit of p^3 - p^2 to be 0, the units digit of p^3 and p^2 must be the same. Thus the units digit of p can be 0, 1, 5 or 6.

Intersection of values is 6, thus the units digit of p + 3 is 6 + 3 = 9.

Answer: B.
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Re: Given that p is a positive even integer with a positive [#permalink] New post 25 Feb 2014, 21:09
Argh, stung again by not reading the question correctly.

Thought it could have been 5 or 6, so picked e.

Until I reviewed the answer and re-read them q stem.
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Re: Given that p is a positive even integer with a positive [#permalink] New post 04 Mar 2014, 20:32
actionj wrote:
Argh, stung again by not reading the question correctly.

Thought it could have been 5 or 6, so picked e.

Until I reviewed the answer and re-read them q stem.



As stated above; numbers ending with 0, 1 , 5 , 6 have squares, cubes or any high power returning the same units digit at the end

Example:

5 ^ 10000 would also have 5 at the end
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Re: Given that p is a positive even integer with a positive   [#permalink] 04 Mar 2014, 20:32
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