NoHalfMeasures wrote:
a is a nonzero integer. Is a^a greater than 1?
(1) a < -1
(2) a is even
Target question: Is a^a greater than 1? Given: a is a nonzero integer. Statement 1: a < -1 Let's start TESTING some values of a and see if we
discover a PATTERNa = -2. Here a^a = (-2)^(-2) = 1/(-2)^2 = 1/4
a = -3. Here a^a = (-3)^(-3) = 1/(-3)^3 = 1/(-27) = - 1/27
a = -4. Here a^a = (-4)^(-4) = 1/(-4)^4 = 1/256
a = -5. Here a^a = (-5)^(-5) = 1/(-5)^5 = 1/(some negative value) = something negative
a = -6. Here a^a = (-6)^(-6) = 1/(-6)^6 = 1/(some positive integer) = a positive fraction that's less than 1
a = -7. Here a^a = (-7)^(-7) = 1/(-7)^7 = 1/(some negative value) = something negative
.
.
.
As we can see, when a is a NEGATIVE
EVEN number,
a^a = some positive fraction that's LESS THAN 1When a is a NEGATIVE
ODD number,
a^a = some negative value that's LESS THAN 1So, in both possible cases,
a^a is definitely LESS THAN 1Since we can answer the
target question with certainty, statement 1 is SUFFICIENT
Statement 2: a is even We already saw in statement 1 that if a is a
NEGATIVE EVEN integer, then
a^a = some positive fraction that's LESS THAN 1What if a is a
POSITIVE EVEN integer?
Let's test some possible values:
a = 2. Here a^a = (2)^(2) = 4. Here,
a^a is GREATER THAN 1So, we'll stop right here.
Since we cannot answer the
target question with certainty, statement 2 is NOT SUFFICIENT
Answer: A
Cheers,
Brent
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