question "Terminator PQ"

[dl]

my question is how is this question even able to be solved? i feel that there is not enough information, also, the explanation doesnt explain why 2^X * 5^Y is valid for all cases. here is the question and the 2 statements:


If a, b, c, d and e are integers and p = 2^a3^b and q = 2^c3^d5^e, is p/q a terminating decimal?

(1) a > c
(2) b > d

i feel they need to give some information about e or why the solution states, "(Any integer divided by a power of 2 or 5 will result in a terminating decimal.)"

for example,

a=4
b=3
c=2
d=1
e=5

this satisfies both statments, therefore you have (2^4*3^3)/(2^2*3^1*5^5)
this simplifies to (2^2*3^2)/(5^5)
which is then (4*9)/(5^5)

and that is definitely not a terminating decimal, or in any case, your denominator will be 5^x, and it cant be determined whether that is a terminating decimal.


Thanks

[RonPurewal]

my question is how is this question even able to be solved? i feel that there is not enough information, also, the explanation doesnt explain why 2^X * 5^Y is valid for all cases. here is the question and the 2 statements:


If a, b, c, d and e are integers and p = 2^a3^b and q = 2^c3^d5^e, is p/q a terminating decimal?

(1) a > c
(2) b > d

i feel they need to give some information about e or why the solution states, "(Any integer divided by a power of 2 or 5 will result in a terminating decimal.)"

for example,

a=4
b=3
c=2
d=1
e=5

this satisfies both statments, therefore you have (2^4*3^3)/(2^2*3^1*5^5)
this simplifies to (2^2*3^2)/(5^5)
which is then (4*9)/(5^5)

and that is definitely not a terminating decimal, or in any case, your denominator will be 5^x, and it cant be determined whether that is a terminating decimal.


Thanks

sure that's a terminating decimal. it's 36/3125 = 0.01152, which terminates.

the rule, which i'm not sure whether you understand completely, is this:
AFTER YOU REDUCE A FRACTION, the decimal equivalent will terminate if the denominator contains no primes other than 2's and 5's.
the converse is also true: if there is any prime other than 2 or 5 in the denominator, then the decimal will be repeating (i.e., it will not terminate).

if by "information" you mean proof, the basic sketch of the proof is this: when you perform long division to create a decimal, once you get past the decimal point, you're basically performing the equivalent of dividing 10, or 100, or 1000, ..., by the denominator.
the only primes that will divide evenly into those numbers are 2 and 5, and the only way to get the decimal to terminate is to get the denominator to divide evenly into one of these powers of ten. that's why it works.
the full proof is pretty complicated, but it doesn't matter; obviously you aren't going to use the proof on the gmat. you're just going to use the result; see the boldface part above.