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02 Oct 2024, 11:04
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Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
(N/A)
Question Stats:
75% (00:40) correct
25% (01:36) wrong based on 4 sessions
History
Date
Time
Result
Not Attempted Yet
r and t are consecutive integers and \(p = r^2 + t.\)
Quantity A
\((-1)^p\)
Quantity B
-1
Quantity A
\((-1)^p\)
Quantity B
-1
03 Mar 2025, 09:02
Kudos
Bookmarks
OE
Recall that
\((-1)^p =\) { 1 if p is an even integer and -1 if p is an odd integer}
Since \(p = r^2 + t,\) the value of (-1)p depends on whether \(r^2 + t\) is odd or even.
If r is an odd integer, then \(r^2\) is an odd integer and, since r and t are consecutive integers, t is an even integer. In this case, p is the sum of an odd integer and an even integer and is therefore an odd integer.
Similarly, if r is an even integer, then \(r^2\) is an even integer and t is an odd integer. In this case, p is the sum of an even integer and an odd integer and is therefore an odd integer.
In both cases, p is an odd integer. It follows that \((-1)^p = -1,\) and the correct answer is Choice C.
Recall that
\((-1)^p =\) { 1 if p is an even integer and -1 if p is an odd integer}
Since \(p = r^2 + t,\) the value of (-1)p depends on whether \(r^2 + t\) is odd or even.
If r is an odd integer, then \(r^2\) is an odd integer and, since r and t are consecutive integers, t is an even integer. In this case, p is the sum of an odd integer and an even integer and is therefore an odd integer.
Similarly, if r is an even integer, then \(r^2\) is an even integer and t is an odd integer. In this case, p is the sum of an even integer and an odd integer and is therefore an odd integer.
In both cases, p is an odd integer. It follows that \((-1)^p = -1,\) and the correct answer is Choice C.
