Given that a, b, c, and, d are non-negative integers, is the

Question

Given that a, b, c, and, d are non-negative integers, is the fraction  \frac{(ad)}{(2^a3^b4^c5^d)}  a terminating decimal?

(1)  d = \frac{(1 + a)(a^2 – 2a + 1)}{((a – 1)(a^2 – 1))}

(2)  b = (1 + a)(a^2 – 2a + 1) – (a – 1)(a^2 – 1)

arps · Answer

IMO B, what is OA?

explanation:

1) gives ad= a/ 2........ not sufficient

2) gives, b=0, for the expression tobe a terminating decimal, power of 3=0, as 1/3= 0.333333..... non terminating, 1/2, 1/4, 1/5 are all terminating decimals

gskang · Answer

Could you please confirm if (a2-1) is a mistake in statement 1 and it should rather be (a^2-1)?

My calculations are based on my assumption above.

Statement 1 gives us a=d, which means that we will have (2*5)^a = 10^a in the denominator but I believe this is not sufficient as we still have 3^b in the denominator that can possibly create a non-terminating decimal. 
Not sufficient in my opinion.

Statement 2 give us b = 0, which means that there is no component of 3 in the denominator now. Thus there is no chance of getting a non-terminating decimal from rest of the numbers as 2, 4, and 5.
This should be sufficient. Hence answer is B; please correct me if I used any wrong assumptions.

Bunuel · Answer

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tessyoncom · Answer

for a fraction to be a terminating decimal, the denominator must contain  only  factors of 2 and 5. since evaluating the powers of the denominator, statement 2 says b=0 hence 3^b is eliminated and the fraction is left with only 2's and 5's as the denominator. therefore statement 2 is sufficient

MathRevolution · Answer

Forget conventional ways of solving math questions. In DS, Variable approach is

the easiest and quickest way to find the answer without actually solving the

problem. Remember equal number of variables and equations ensures a solution.

Given that a, b, c, and, d are non-negative integers, is the fraction (ad)/(2^a3^b4^c5^d) a terminating decimal?

(1) d = (1 + a)(a^2 – 2a + 1)/((a – 1)(a^2 – 1))
(2) b = (1 + a)(a^2 – 2a + 1) – (a – 1)(a^2 – 1)
 
transforming the original condition and the question by variable approach method, in order for (ad)/(2^a3^b4^c5^d)=terminating decimal we need ad (numerator) to have 3 as a factor. Or, we would need 3^b removed from the denominator. This is because the terminating decimal can only have 2 or 5 as their prime factors in the denominator

But in case of 2), if b = 0, 3^b=3^0=1 and the denominator is (2^a)(4^c)(5^d). Therefore the prime factor of the denominator is 2 or 5. Since the condition is sufficient. the answer is B.

Psiva00734 · Answer

if any fraction it has to be terminating decimal. 
denominator can have 2 or 5 or 2 and 5. it can terminate decimal.
whatever powers in 2 and 5
In question stem mentions a, b, c and d are non negative integers. It means it can be 0 and +ve integers.

In fraction ad/2^a 3 ^ b 4^c 5^d.
we can 4^c as 2 ^ 2c.
Now 2, 4,5 powers are terminating decimals. In 4 if power c value is 0 it becomes 1.
fraction to terminate ad/3^ b must be an integer. not a fraction.

in statement 1
evaluate d=1. we have to know value of b. so not sufficient.

In statement 2. 
evaluate b=0.
so 3powerB 3^0 =1.
so ad/1is integer. 
so option B is sufficient.

.

dina98 · Answer

I'm slightly confused. If a or d = 0, is ad / (10)^n = terminating decimal? Is 0 a terminating decimal?

MrJglass · Answer

Can someone please solve the second statement and show the working?
b=(1+a)(a^2 -2a + 1) - (a-1)(a^2 -1 )
Thanks!

amanvermagmat · Answer

Hello

Given that b = (1 + a)(a^2 – 2a + 1) – (a – 1)(a^2 – 1)
Now, a^2 - 2a + 1 is square of (a-1) and can thus be written as (a-1)^2 or (a-1)(a-1). 
a^2 - 1 can be read as a^2 - 1^2 and can thus be written as (a+1)(a-1)

So now we can rewrite b as:
b = (a+1)(a-1)(a-1) - (a-1)(a+1)(a-1) 
As you can see, both terms are same and thus their difference will be 0.

MrJglass · Answer

Awww! I see. Thanks a lot, I'm really grateful. 
What a relief!

fireagablast · Answer

Was confused because the theory wasn't shared.
In order for something to be a terminating decimal, it needs to be divided by ONLY prime factors 2 and 5.
Example:

2/5 -> terminates
2/2 -> terminates
2/(5*2) -> terminates
2/4 -> terminates
2/7 -> doesn't terminate

Now the problem:
We have (ad) divided by several prime factors to different exponents. The exponents largely don't matter, but it IS important to notice that we have a "3". This "3" throws off the terminating decimal requirement.

(1) this simplifies to d=1. This doesn't help because we still have a "3" in the denominator. B could be 0 which would make it terminating, but b is currently infinity. NS
(2) this simplifies to b=0. This means 3^b = 1. now we have only 2*1*4*5 in the denominator. This fits the condition for a terminating decimal. SUFF

Ans. B

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