Any decimal that has only a finite number of nonzero digits

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

Bunuel · Accepted Answer

Theory: 
Reduced fraction  \frac{a}{b}  (meaning that fraction is already reduced to its lowest term) can be expressed as terminating decimal  if and only   b  (denominator) is of the form  2^n5^m , where  m  and  n  are non-negative integers. For example:  \frac{7}{250}  is a terminating decimal  0.028 , as  250  (denominator) equals to  2*5^2 . Fraction  \frac{3}{30}  is also a terminating decimal, as  \frac{3}{30}=\frac{1}{10}  and denominator  10=2*5 .

Note that if denominator already has only 2-s and/or 5-s then it doesn't matter whether the fraction is reduced or not.

For example  \frac{x}{2^n5^m} , (where x, n and m are integers) will always be the terminating decimal.

We need reducing in case when we have the prime in denominator other then 2 or 5 to see whether it could be reduced. For example fraction  \frac{6}{15}  has 3 as prime in denominator and we need to know if it can be reduced.

Hope it helps.

jeeteshsingh · Answer

IMO B..

Explanation:

\frac{p}{q} = \frac{2^a*3^b}{2^c*3^d*5^e}

For any fraction to be terminating... the denominator should be in form of  2^m*5^n  in  it's lowest form  where m and n are  non negative integers - could be 0 also...

1. a > c
Therefore a-c (let say k) > 0

Hence:
 \frac{p}{q} = \frac{2^k*3^b}{3^d*5^e} ... which is could be a terminating decimal if b>d... or non terminating... if b<d. INSUFF...

For example the fraction could be ....  \frac{4*3}{2*9*5}=0.13333...  or  \frac{4*9}{2*3*5}=0.12

2. b>d
Therefore b-c (let say n) > 0
Hence:
 \frac{p}{q} = \frac{2^a*3^n}{2^c*5^e} ... This is clearly a terminating decimal as the denominator would be in a form of  2^m*5^n

For example the fraction could be ....  \frac{4*3}{2*5}=0.12  or  \frac{4*3}{8*5}=0.3

BarneyStinson · Answer

Interesting question and what nitish suggested was a good formula but when you apply it, the answer should actually be different.

The answer should be A. Only statement 1 is sufficient to say that the ratio p/q is non-terminating definitively.

Stmt-1 deals with the relationship between a and c, so we know clearly that 2 will not be there in the denominator.

Stmt-2 on the other hand relates b with d, so we don't know if 2 will be there in the denominator or not.

2^a/2^c = 2^(a - c) when a > c in the numerator and will be 1/2^(c - a) when a < c.

cipher · Answer

Can you explain why A is the answer?

I guess A only tells us that 2 wont be there in the denominator but does not tell us anything about 3, and now if 3 will be there in the denominator it will be non terminating decimal but if 3 wont be there then it will be a terminating decimal and hence its not sufficient

on the other hand st 2 clearly tells that 3 wont be there in the denominator and hence its sufficient.

Correct me if I am missing some thing here ..!

alexBLR · Answer

I'm not Nitish, but if I can answer your question, then yes, it is a rule. For the number to be a terminating decimal in denominator it has to have 2^x*5^y and x or y can be 0

Spidy001 · Answer

1. Not sufficient 
 As 3 will be there in denominator or not.

2.sufficient 
 As we know 3 is not in the denominator and denominator has a 5.

Answer is B.

Posted from my mobile device

vibhav · Answer

picked B. knew that 2/5 in the denominator will lead to a terminating decimal irrespective of a numerator. However didnt know of the formal rule. very helpful.

cyberjadugar · Answer

Hi,

p = 2^a3^b  and  q = 2^c3^d5^e 
 p/q = 2^(a-c)3^(b-d)5^(-e)

division by 2 or 5 (raised to any power will result to terminating decimal)

only check is on b-d

so, using (2)
b-d > 0,
thus 3 is not in denominator, hence p/q is terminating decimal.

Answer is (B).

Regards,

Rice · Answer

Bunuel,

if denominator has only 2 or only 5, it seems that the fraction will also be a terminating decimal. Right?

gmatsaga · Answer

My answer is B. What's the OA?

The only determinant if p/q is a terminating decimal is 3^b/3^d since a fraction with the format a/b is terminal if b is a power of 2 or 5 or both.

Bunuel · Answer

Yes.

Reduced fraction  \frac{a}{b}  (meaning that fraction is already reduced to its lowest term) can be expressed as terminating decimal  if and only   b  (denominator) is of the form  2^n5^m , where  m  and  n  are  non-negative integers .

Notice that when n or m equals to zero then the denominator will have only 2's or 5's.

Hope it's clear.

Bunuel · Answer

OA is given in the initial post and it's B. 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 Check this: any-decimal-that-has-only-a-finite-number-of-nonzero-digits-90504.html#p689656 So, according to that \frac{p}{q}=\frac{2^a*3^b}{2^c*3^d*5^e} will be terminating decimal if 3^d in the denominator can be reduced, which will happen when the power of 3 in the numerator is more than or equal to the power of 3 in the denominator, so when b\geq{d} . As we can see (1) is completely irrelevant to answer whether b\geq{d} , while (2) directly answers the question by stating that b>d . Answer: B. Hope it's clear.

pavanpuneet · Answer

In the given problem, if a and c are not considered, then there may be a case when 2^a can be completely divided by 2^c, in that case, how the answer can be b? or if a-c is positive. Please let me know where am I thinking wrong?

Bunuel · Answer

Not sure I understand what you mean above. What difference does it make whether 2^a is reduced or not? Or whether a-c is positive? Please read this: any-decimal-that-has-only-a-finite-number-of-nonzero-digits-90504.html#p689656 and this: any-decimal-that-has-only-a-finite-number-of-nonzero-digits-90504.html#p1096496 for theory. Solution: any-decimal-that-has-only-a-finite-number-of-nonzero-digits-90504.html#p1096500

dvinoth86 · Answer

If the denominator in a fraction has only 2 or/and 5 --> terminating decimal
If the denominator has any other prime factor other than 2 or 5 --> non-terminating decimal.

Hence is this question, we need to find out if 3 will be present in the denominator or not! . which means we need to find out if b > d . 
Hence OA : B.

hermit84 · Answer

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

A/B will be terminating(T) only if 1/B is T. 
B = 1/5^e will be always T , in the same way as 1/3^e will always be non- terminating(NT).
Product of a NT with a T will always be NT.

In the numerator of p/q, powers of 2 and 3 can be +ve or -ve. Power of 2 doesn't have any affect on the T behaviour of p/q, but if power of 3 is -ve, it will go down to denominator and 1/3^x is always NT and make p/q NT.

stmt 1: a > c means power of 2 will be positive - INSUFFICIENT.
stmt 2: b > d means power of 3 will be positive and 3^(b-d) will remain in the numerator. Thus, p/q will be T. SUFFICIENT

Answer is B

lpetroski · Answer

Brunel -

Just to be 100% clear -- ANY fraction must have a denominator of 2, 5, or some multiple of the two numbers to terminate?

ENGRTOMBA2018 · Answer

Yes. Fractions with denominators multiples of  2 or 5 or 2 AND 5 ONLY  will give you a terminating decimal representation.

Examples. x/25, x/100, x/125, x/200 etc.

ajdse22 · Answer

My approach to solve this problem:
p/q = (2^(a-c)*3^(b-d))/5^e
statement 1) a>c.
=> the 2^(a-c) is an integer, more so , since 2 is positive, it will return a positive integer.
case1) b>d then 3^(b-d) is an integer . Now whatever be the value of e (again integer) we will have a denominator with powers of 5. We know that if we divide by 5 or its powers, we always get a terminating decimal. The reason is that we are always multiplying 5 with itself. To see this logic. let us say we are dividing a number by 5^2. Simply divide the number by the first 5. This always gives a terminating decimal. Then divide that terminating decimal number with another 5 remaining in the denominator. This logic can be extended to any powers of 5.

case 2) let us try to reduce this to 2/3 ..it is possible to set b-d = 1 and e=0 and a-c=1. We know 2/3 is non terminating
Therefore insuff

statement #2
b>d:- This guarentees that 3^(b-d) is an integer.
Now if a> or =c and e>0 this implies that we will divide an integer by 5 .. answer is terminating decimal
now if a <c and e >0 this implies that we divide integer by 10, answer is yes
what if e< 0 then either we will have p/q as integer or a number divisible by 2 .. we know this will also be a terminating decminal

So statement #2 is sufficient -
B

Any decimal that has only a finite number of nonzero digits

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Any decimal that has only a finite number of nonzero digits

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