ziyuenlau wrote:
If positive even integer p has a positive units digit and the units digit of (\(p^{3} – p^{2}\)\(\)) is equal to 0, what is the units digit of the quantity \(p + 3\)?
A. 3
B. 6
C. 7
D. 9
E. It cannot be determined from the given information.
Official answer from
Manhattan Prep.
The variable p is a positive even integer and the units digit of this number is also positive. Therefore, the units digit has four possible values: 2, 4, 6, or 8. (O is not a possibility, since 0 is not positive.)
This is useful information because the units digit of a number is the only value that determines the new units digit when that number is squared or cubed. For instance \(1,394^{2} =\) some very large number whose units digit is 6, because \(4^{2} = 16\).
In other words, only the units digit of \(p\) will determine the units digit of \(p^{3}\) and \(p^{2}\) . Further, since the units digit of \(p^{3} – p^{2} = 0\), you’re looking for a circumstance in which the square of the units digit is equal to the cube of the units digit.
When does this occur? Test the possible units digits 2, 4, 6, and 8 to see.
Only when \(p = 6\) are the units digits of \(p^{3}\) and \(p^{2}\) equal—and therefore the units digit of \(p^{3} – p^{2} = 0\). As a result, the units digit of p must equal 6, and the units digit of \(p + 3\) must equal 9.
The correct answer is (D).
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