Data Sufficiency

Here is a question:

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 non-negative integers and p = 2^a*3^b and q = 2^c*3^d*5^e, is p/q a terminating decimal?

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

Can you please provide an answer to this?

The solution provided by Ian is perfect and very informative, but I guess many test takers go to the exam without this knowledge... that´s why I believe the solution I present below is also nice:

From the "a, b, c, d and e are non-negative integers" hypothesis, I believe the question should be "seen" as:

2^(a-c) * 3^(b-d) over 5^e is a terminating decimal ?

Important: this is just a visually helpful thing, because (for instance) the value of (a-c) may be negative, and that means that 2^(a-c) can be at the denominator, "in reality"...

(1) This sttm tells us that 2 is really in numerator, but what about the 3(´s) ?

> Take a = 2, c =1 (to be only 2^1) and b = 2 and d = 1 (to let 3 be 3^1, so numerator...) and e = 1 , then we have: (2 * 3) over 5, and if you multiply both numerator and denumerator by 2 , you get (2^2 * 3) over... 10, that is, certainly terminating because it is an integer divided by 10, so you just move decimal point, you do not "alter terminallity"...

> Take a =2 , c = 1 (to be only 2^1 again) and b = 1 and d = 2 (to let 3 be 3^(-1), so denumerator) and e =1 , then we have: 2 over (3 times 5) and now we know we are with a non-terminating decimal because of this (for instance):

2/(15) = 2*2 / (15*2) = 4/30 = (1/10) * (4/3) and divide by 10 does not alter the "terminallity of a certain decimal" (as mentioned above) and we know that 4/3 is not terminating, because it is equal to 1+ 1/3 and (1 is an integer and) 1/3 is non-terminating, for sure (0.333333...)

Obs.: this is not stupid calculations, I believe. This is the "insight" that is behind Ian´s statements...

(2) Now we know that 3´s are on the numerator, therefore we may have only 2´s , only 5´s or both in the denumerator. From all shown, you should be able to recognize that 2´s and 5´s are no problem, because multiplying by 10´s in enough quantity you turn the fractions into integers, therefore terminating decimals.

This one DECIDES affirmatively on the question asked, that is, (2) is sufficient.

Regards,
Fabio.

The ans is B

when we simplify the question we get 2^(a-c) * 3^(b-d) / 5^e

when we have prime factors other than 2 and 5 in the denominator we get non-terminating decimal.

1) a>c is in sufficient. If a>c or a<c it doesnt matter since we don't know about b and d.
2) if b>d there is no 3 in the denominator. So, we have terminating number.

Hence 2 alone is sufficient to ans the question.

mgmat cat question hmmmmmmmmm

already enough on this by the experts, my say in short.

remember 2,4,6,8 and 10(because it has 2 and 5) are terminators(the bad guys)(schwarzenegger in terminator 1)


any fraction with additional 3,6(because it has 3),7 and 9(because it has 3) as denominators is not a terminator and the good guy(schwarzenegger in terminator 2)

hope this helps, u'll remember this now

this question can simply be rephrased as

is the no of 3's in p greater than no of 3's in q ???
i'e
is b>d

the answer is right in one of the statements 8)

Nothing to add here. +1 for B

IMO:B

Waaaaaaaaaaaaaay to hard

p/q = 2^(a-c)*3^(b-d)/5^e

Except b<d where 3 comes in the denominator in all other cases the fraction is a terminating decimal

Statement 1: INSUFFICIENT

Statement 2: SUFFICIENT

if the fraction has powers of 2 or 5 only hen the fraction is terminating decimal
S1: a>c does not tell about power of 3
S2: b-d>0 hence the fraction is terminating decimal
(B) is ans

Ian Stewart wrote:

chrisjim5 wrote:Here is a question:

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 non-negative integers and p = 2^a*3^b and q = 2^c*3^d*5^e, is p/q a terminating decimal?

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

Can you please provide an answer to this?

You can recognize whether a fraction will produce a terminating decimal by:

1. Reducing your fraction completely

2. Then looking only at the prime factors of the denominator. If the denominator has any prime factor besides 2 or 5, the fraction will give a *repeating* (infinite) decimal. If the only prime factors of the denominator are 2 and/or 5, the fraction will give a terminating decimal.

So fractions like 3/16 (only prime factor of denominator is 2), 9/125 (only prime factor of denominator is 5) and 3/40 (only prime factors of denominator are 2 and 5) will all produce terminating decimals. Fractions like 1/13, 9/35, and 11/120 will all produce non-terminating decimals since each is completely reduced, and has a factor different from 2 or 5 in the denominator. The first step above is critical; while a fraction like 7/35 might appear to have a factor of 7 in the denominator, that 7 actually cancels with the 7 in the numerator to give us 1/5, a terminating decimal.

So in this question, we have the fraction:

(2^a*3^b) / (2^c*3^d*5^e)

The only reason this might not terminate is because of the 3's; if our 3^d in the denominator does not cancel out completely, we will get a repeating decimal. If it does cancel, we will get a terminating decimal. Statement 2 tells us that it will cancel completely, so is sufficient. Statement 1 doesn't help. So the answer is B.

Nice Logic Ian.Thanks!!
:arrow: If the denominator has any prime factor besides 2 or 5, the fraction will give a *repeating* (infinite) decimal. If the only prime factors of the denominator are 2 and/or 5, the fraction will give a terminating decimal.

Can someone provide and easier to understand explanation please?

Rastis wrote:Can someone provide and easier to understand explanation please?

Relatively easier approach for you.

Lets put this question as 2^(a-c).3^(b-d)/5^e.

Its for sure that this question wishes us to predict nature of 2^(a-c).3^(b-d)/5^e by deducing nature of 2^(a-c), 3^(b-d), & 5^e.

=> lets put a=c=b=d=0. This will help us find the nature of 5^e.

=> so 2^(a-c).3^(b-d)/5^e = 2^0.3^0/5^e = 1/5^e.

=> Now you may try few values of e such as 1, 2, 3... to see if 1/5^e is terminating or non-terminating. We can conclude 1/5^e is terminating.So 5^e has no role to decide if expression is non-terminating.

Now lets put b=d=e=0. This will help us find the nature of 2^(a-c).

=> so 2^(a-c).3^(b-d)/5^e = 2^(a-c).3^0/5^0 = 2^(a-c).

=> Now you may try few values of (a-c) such as ...., -3, -2, -1 to see if 2^(a-c) is terminating or non-terminating. We can conclude 2^(a-c) is terminating.So 2^(a-c) has no role to decide if expression is non-terminating. Whether a > < c. So stat. 1 is not necessary. Redundant.

Now lets put a=c=e=0. This will help us find the nature of 3^(b-d).

=> so 2^(a-c).3^(b-d)/5^e = 2^0.3^(b-d)/5^0 = 3^(b-d).

=> Now you may try few values of(b-d) such as ...., -3, -2, -1 to see if 3^(b-d) is terminating or non-terminating. We can conclude 3^(b-d) is non-terminating if (b-d) is negative.So to make it terminating (b-d) should be positive or b>d. So stat. 2 is not necessary & sufficient.


Ans B.

Rastis wrote:Can someone provide and easier to understand explanation please?

My explanation is similar to the ones above, but perhaps you will find it helpful:

https://www.beatthegmat.com/terminating- ... 92476.html

B in answer.