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Given that a, b, c, and, d are non-negative integers, is the  [#permalink]

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Question Stats: 61% (02:26) correct 39% (02:20) wrong based on 304 sessions

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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)

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Originally posted by enigma123 on 09 Jan 2012, 22:01.
Last edited by Bunuel on 12 Aug 2014, 06:45, edited 2 times in total.
Renamed the topic, edited the question and added the OA.
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Given that a, b, c, and, d are non-negative integers, is the  [#permalink]

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1
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.
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Re: Given that a, b, c, and, d are non-negative integers, is the  [#permalink]

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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 option B is sufficient.

.
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Re: Given that a, b, c, and, d are non-negative integers, is the  [#permalink]

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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
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Re: Given that a, b, c, and, d are non-negative integers, is the  [#permalink]

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enigma123 wrote:
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) (a2 – 1)
(2) b = (1 + a) (a^2 – 2a + 1) – (a – 1) (a^2 – 1)

Guys - any idea what will be the correct answer and also what is the concept behind solving this?

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.
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Re: Given that a, b, c, and, d are non-negative integers, is the  [#permalink]

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enigma123 wrote:
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)

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Re: Given that a, b, c, and, d are non-negative integers, is the  [#permalink]

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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
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Re: Given that a, b, c, and, d are non-negative integers, is the  [#permalink]

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I'm slightly confused. If a or d = 0, is ad / (10)^n = terminating decimal? Is 0 a terminating decimal?
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Given that a, b, c, and, d are non-negative integers, is the  [#permalink]

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Can someone please solve the second statement and show the working?
b=(1+a)(a^2 -2a + 1) - (a-1)(a^2 -1 )
Thanks!
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Re: Given that a, b, c, and, d are non-negative integers, is the  [#permalink]

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

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.
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Given that a, b, c, and, d are non-negative integers, is the  [#permalink]

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

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.

Awww! I see. Thanks a lot, I'm really grateful.
What a relief!
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Re: Given that a, b, c, and, d are non-negative integers, is the  [#permalink]

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