Any decimal that has only a finite number of nonzero digits

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.

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 ..!

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

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.

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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.

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,

Bunuel,

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

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.

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.

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.

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?

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

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.

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

Brunel -

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

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.

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