“
x
+
y
is divisible by 6” only.
To solve this problem with examples, make
a short list of possibilities for each of
x
and
y
:
x
= 18, 36, 54…
y
= 12, 24, 36…
Now try to
disprove
the statements by trying several combinations of
x
and
y
above. In the first statement,
x
+
y
could be 18 + 12 = 30, 54 + 12 = 66, 36 +
24 = 60, or many other combinations. All of those combinations are multiples
of 6. This makes sense, as
x
and
y
individually are multiples of 6, so their sum
is, too. The first statement is true.
To test the second statement,
xy
could be 18(12) = 216, which is
not
divisible
by 48. Eliminate the second statement.
As for the third statement,
could be
, which is not even an integer (and
therefore not divisible by 6), so the third statement is not necessarily true.
19.
(D).
This problem is most easily solved with an example. If
p
= 7 and
q
=
6, then
pq
= 42, which has the factors 1 & 42, 2 & 21, 3 & 14, and 6 & 7.
That’s 8 factors, but read carefully! The question asks how many factors
greater than 1
, so the answer is 7. Note that choosing the smallest possible
examples (
p
= 7 and
q
= 6) was the right move here, since the question asks
“at least how many factors … ?” If testing
p
= 70 and
q
= 36, many, many
more factors would have resulted. The question asks for the minimum.
20.
(C).
This problem is most easily solved with an example. If
r
= 10 and
s
=
9, then
rs
= 90. The positive factors of 90 are 1 & 90, 2 & 45, 3 & 30, 5 & 18,
6 & 15, and 9 & 10. Count to get a minimum of 12 positive factors.
21.
(D).
If
t
is divisible by 12, then
t
2
must be divisible by 144 or 2 × 2 × 2 ×
2 × 3 × 3. Therefore,
t
2
can be divided evenly by 2 at least four times, so
a
must be at least 5 before
might not be an integer.
Alternatively, test values. If
t
= 12,
. Plug in the choices as
possible
a
values, starting with the smallest choice and working up:
(A)
Since
= 36, eliminate.
(B)
Since
= 18, eliminate.
(C)
Since
= 9, eliminate.
(D)
= 4.5. The first choice for which
might not be an integer
is (D).
22.
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