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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.

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 _________________

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. _________________

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..

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

Re: Given that p is a positive even integer with a positive [#permalink]

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28 Sep 2013, 12:31

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Re: 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.

Re: Given that p is a positive even integer with a positive [#permalink]

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03 Oct 2015, 20:59

Hello from the GMAT Club BumpBot!

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