Terminator PQ Question

[mhweiler]

For the following problem, manhattan GMAT's explanation says that and integer divided by 2^x5^y will be terminating if x and y are non-negative integers. How can we tell that x, and y (in this case c and e) are non-negative? All we have from the problem is that they are integers...

Any decimal that has only a finite number of nonzero digits is a terminating decimal. For example, 36, 0.72, and 3.005 are terminating decimals.
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

MANHATTAN GMAT EXPLANATION:

For fraction p/q to be a terminating decimal, the numerator must be an integer and the denominator must be an integer that can be expressed in the form of 2x5y where x and y are nonnegative integers. (Any integer divided by a power of 2 or 5 will result in a terminating decimal.)
The numerator p, 2a3b, is definitely an integer since a and b are defined as integers in the question.

The denominator q, 2c3d5e, could be rewritten in the form of 2x5y if we could somehow eliminate the expression 3d. This could happen if the power of 3 in the numerator (b) is greater than the power of 3 in the denominator (d), thereby canceling out the expression 3d. Thus, we could rephrase this question as, is b > d?

(1) INSUFFICIENT. This does not answer the rephrased question "is b > d"? The denominator q is not in the form of 2x5y so we cannot determine whether or not p/q will be a terminating decimal.

(2) SUFFICIENT. This answers the question "is b > d?"

The correct answer is B.

[RonPurewal]

For the following problem, manhattan GMAT's explanation says that and integer divided by 2^x5^y will be terminating if x and y are non-negative integers. How can we tell that x, and y (in this case c and e) are non-negative? All we have from the problem is that they are integers...

heh. good catch.
you have a valid point, but, fortunately, it turns out not to change the answer to this particular problem.

here's the story:
if you have ZERO or NEGATIVE exponents in the denominator, then those don't create fractions at all. in fact, dividing by an 0 or negative power of an integer is equivalent to multiplying by an integer.

as an illustration it should be good enough for me just to show you one example.

if you have, say, 5^-2 in a denominator, then you're dividing by 1/25. that's the same as multiplying by 25, so, in effect, that wouldn't be a "denominator" at all.
IE fractions with negative powers in their denominators aren't actually fractions in the first place; they're just integers written in really weird ways.

--

on the other hand:

in a real gmat problem, you will, indeed, never be faced with the prospect of negative exponents in a prime factorization. it just won't happen (although zero exponents are a possibility).

i'll submit this problem for that minor edit (i.e., it should stipulate at least that the exponents are nonnegative, although, as stated above, that doesn't actually affect the outcome of this problem).

[shakil]

Would the test creators expect that we memorize the fact that 2 and 5 and the magic number's that yield this special rule? -- or is there some obvious way we're supposed to derive this on the fly?

[esledge]

Yes, it would be fair game for the GMAT to expect you to see that 2's and 5's are "magic numbers." You might find this related thread interesting--see page 2, where I showed an "on the fly" type derivation.

for-fraction-p-q-to-be-a-terminating-decimal-the-numerator-t741-15.html

[navdeep_bajwa]

MGMAT explanation The numerator p, 2^a*3^b, is definitely an integer since a and b are defined as integers in the question.

how p can be integer what if a or b or both are negative

[manochsa]

The explanation to this problem states "Any integer divided by a POWER of 2 or 5 will result in a terminating decimal".

I've read elsewhere that any integer divided by a MULTIPLE of 2 or 5 will result in a terminating decimal.

Which must it be -- a power or a multiple?

Thanks,

Sanjay

[dmahler]

It's a power of 2 or 5, not a multiple.

Take the case of 15, which is a multiple of 5. 1/15 yields a non-terminating decimal - .066666...

[crutchfield.adam]

Something doesn't seem right here, even though this question has been revised to have a, b, c, d, and e as nonnegative integers rather than just integers. If c=e=0, then the denominator would just be 1 and would not contain any 2s or 5s. With the possibility that c=e=0, isn't it true that the information given in (2) that b>d could mean that p/q is either a terminating decimal or is not a terminating decimal, making for a possible "yes" and possible "no" answer to the question? If so, this would lead to (E) as the correct answer.

[RonPurewal]

hmm

Something doesn't seem right here, even though this question has been revised to have a, b, c, d, and e as nonnegative integers rather than just integers. If c=e=0, then the denominator would just be 1 and would not contain any 2s or 5s.

two things quite wrong here.

first, your condition c=e=0 isn't enough to guarantee that the denominator is 1, since there is still that pesky 3^d in there.

second -- if you did have a denominator of 1, then you would wind up with an integer overall. all integers are terminating, since they don't even have decimal places at all!

With the possibility that c=e=0, isn't it true that the information given in (2) that b>d could mean that p/q is either a terminating decimal or is not a terminating decimal, making for a possible "yes" and possible "no" answer to the question?

nope.

with your condition c=e=0, the only quantity left in the denominator is 3^d. if b is greater than d, then the 3^b in the numerator is large enough to swallow the 3^d in the denominator. if that happens, then there's no denominator left at all, so again you have an integer -- which is automatically terminating.

for the future, a tip: when you make these sorts of arguments, you should try to find actual numbers that will corroborate your argument. if you try to do so here, then you won't find such numbers, a situation that may convince you that your initial insight was incorrect.