18
3
, 30
3
, and 222 only.
Because 64 = 2
6
, multiples of 64 would have at
least six 2’s among their prime factors.
Since 12 (which is 2 × 2 × 3) has two 2’s already, a number that could be
multiplied by 12 to generate a multiple of 64 would need to have, at
minimum, the
other
four 2’s needed to generate a multiple of 64.
Since you want the choices that don’t multiply with 12 to generate a multiple
of 64, select only the choices that have
fewer than four 2’s
within their prime
factors.
32.
(D).
When a non-multiple of 3 is divided by 3, the quotient does not
terminate (for instance, = 0.333…).
Since
does
not
repeat forever,
x
must be large enough to cancel out the
3
5
in the denominator. Thus,
x
must be at least 5. Note that the question asks
what
must
be true. Choice (D) must be true. Choice (E),
x
= 5, represents one
value that would work, but this choice does not
have
to be true.
33.
(B).
Since a prime number has only two factors, 1 and itself, (2
a
)(3
b
)(5
c
)
cannot be prime unless the digits
a, b
, and
c
are such that two of the digits are
0 and the third is 1. For instance, (2
0
)(3
1
)(5
0
) = (1)(3)(1) = 3 is prime. Thus,
the only three values of
abc
that would result in a prime number &(
abc
)& are
100, 010, and 001. However, only one of those three numbers (100) is a three-
digit number.
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