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Given that p is a positive even integer with a positive [#permalink]
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08 Aug 2006, 13:59
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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.
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Last edited by Bunuel on 29 Sep 2013, 09:47, edited 1 time in total.
Renamed the topic, edited the question and added the OA.



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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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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?
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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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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
Answer should be D because last digit of p must be +ve and even.
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Re: PS: Digits again [#permalink]
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08 Aug 2006, 14:42
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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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^3p^2 = p^2(p1) = (2k)^2*(2k1)
Since we know that units digit of p^3p^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 2k1=5
=> 2k=6 => p=6 => p+3=9
(D)



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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]
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09 Aug 2006, 17:13
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
Heman



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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 = 6 4 & 4^2=1 6 . Subrtacting units digit does not result in 0.
Try the rest. Only when 6 is in the units digit will 6^3= 21 6 , 6^2=3 6. Subtracting units digit will result in zero.
Hence reqd answer = 6+3 = 9 Choice D
Heman



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yes... zero is neither positive nor negative so you must leave it alone
I just picked p=2, 4, 6 and got p^3p^2= 4, 8, 0
p=6 and p+3=9



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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]
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Given that p is a positive even integer with a positive [#permalink]
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29 Sep 2013, 09:57
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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: D.
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Re: Given that p is a positive even integer with a positive [#permalink]
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25 Feb 2014, 22:09
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 reread them q stem.



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Re: Given that p is a positive even integer with a positive [#permalink]
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04 Mar 2014, 21:32
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 reread 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 endExample: 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]
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03 Oct 2015, 20:59
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Given that p is a positive even integer with a positive [#permalink]
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05 Oct 2015, 00:02
Given: P is an even positive number (i) and units digit of p^3  units digit of p^2= 0 (ii) Required: Units digit of p+3 For (i), The units digit can be 2,4,6 or 8 For (ii), the numbers for which the units digit of cube and square is same are {0, 1, 5, 6} But we know that p is positive and even. Hence p = 6 p+3 = 9 Option D
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Re: Given that p is a positive even integer with a positive [#permalink]
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17 Jun 2016, 03:12
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Bunuel wrote: 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. Apologies for seeing a mistake,shouldn't the answer choice be option D



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Re: Given that p is a positive even integer with a positive [#permalink]
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17 Jun 2016, 04:48
rhine29388 wrote: Bunuel wrote: 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. Apologies for seeing a mistake,shouldn't the answer choice be option D ______________ Edited. Thank you.
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