divisibility problem

remember the following: the PATTERNS in REMAINDER problems will emerge fairly early when you plug in numbers.
there are two kinds of problems that are especially suited to plugging in numbers and looking for patterns:
* digit problems
* remainder problems

pattern recognition is always a good backup strategy, but, on these two problem types, it's pure gold.

therefore, if you don't IMMEDIATELY realize a good theoretical way to do a remainder problem, you should get on the number plugging RIGHT AWAY.

in fact, i got an 800 on this thing, and i would immediately jump to plug-in / pattern recognition on problems such as this one. so don't sweat it if you don't find the theory method.

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statement (1):
just dredge up an exhaustive list of numbers that satisfy this criterion. if division by 2 leaves a remainder of 1, then the number in question must be 1 more than a multiple of 2.
therefore,
1, 3, 5, 7, ...
(you could also figure out this list by pure experimentation)

try these numbers in the prompt question:
x = 1 --> no (the remainder is 1)
x = 3 --> yes

if you don't like dividing numbers that are less than 4 by 4, then just try 5 and 7. 5 gives "no", and 7 gives "yes".

so, insufficient.

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statement (2)
3, 6, 9, 12, ...

3 --> yes
6 --> no
insufficient.

if you don't like dividing numbers that are less than 4 by 4, your first "yes" will occur at x = 15, which is not that far down the list.

so, insufficient.

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together:
take the numbers that appear on both lists. (if you didn't make very long lists - which you certainly didn't have to, since the first 2 numbers of each list are enough to resolve the issue - then you may well have to extend the lists to do this)

those numbers are
3, 9, 15, 21, ...

try them
3 --> yes
9 --> no
still insufficient.

if you don't like dividing numbers that are less than 4 by 4, then just try 9 and 15, which give "no" and "yes" respectively.
still insufficient.

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ans (e)