Author 
Message 
TAGS:

Hide Tags

Senior Manager
Status: Finally Done. Admitted in Kellogg for 2015 intake
Joined: 25 Jun 2011
Posts: 439
Location: United Kingdom
Concentration: International Business, Strategy
GPA: 2.9
WE: Information Technology (Consulting)

Given that a, b, c, and, d are nonnegative integers, is the
[#permalink]
Show Tags
Updated on: 12 Aug 2014, 06:45
Question Stats:
61% (02:25) correct 39% (02:20) wrong based on 301 sessions
HideShow timer Statistics
Given that a, b, c, and, d are nonnegative 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)
Official Answer and Stats are available only to registered users. Register/ Login.
_________________
Best Regards, E.
MGMAT 1 > 530 MGMAT 2> 640 MGMAT 3 > 610 GMAT ==> 730
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.



Manager
Joined: 18 Dec 2011
Posts: 50

Re: Given that a, b, c, and, d are nonnegative integers, is the
[#permalink]
Show Tags
09 Jan 2012, 23:10
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



Intern
Joined: 07 Jan 2012
Posts: 6
Location: United States
WE: Marketing (Other)

Re: Given that a, b, c, and, d are nonnegative integers, is the
[#permalink]
Show Tags
09 Jan 2012, 23:53
enigma123 wrote: Given that a, b, c, and, d are nonnegative 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 (a21) is a mistake in statement 1 and it should rather be (a^21)? 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 nonterminating 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 nonterminating 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.



Math Expert
Joined: 02 Sep 2009
Posts: 59147

Re: Given that a, b, c, and, d are nonnegative integers, is the
[#permalink]
Show Tags
12 Aug 2014, 06:45



Intern
Joined: 04 Mar 2011
Posts: 13

Re: Given that a, b, c, and, d are nonnegative integers, is the
[#permalink]
Show Tags
14 Aug 2014, 17:29
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



Math Revolution GMAT Instructor
Joined: 16 Aug 2015
Posts: 8152
GPA: 3.82

Given that a, b, c, and, d are nonnegative integers, is the
[#permalink]
Show Tags
04 Sep 2015, 04:41
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 nonnegative 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.
_________________
MathRevolution: Finish GMAT Quant Section with 10 minutes to spareThe oneandonly World’s First Variable Approach for DS and IVY Approach for PS with ease, speed and accuracy. "Only $79 for 1 month Online Course""Free Resources30 day online access & Diagnostic Test""Unlimited Access to over 120 free video lessons  try it yourself"



Manager
Status: single
Joined: 19 Jan 2015
Posts: 83
Location: India
GPA: 3.2
WE: Sales (Pharmaceuticals and Biotech)

Re: Given that a, b, c, and, d are nonnegative integers, is the
[#permalink]
Show Tags
06 Sep 2015, 21:30
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.
.



Manager
Joined: 14 Jul 2014
Posts: 158
Location: United States
GMAT 1: 600 Q48 V27 GMAT 2: 720 Q50 V37
GPA: 3.2

Re: Given that a, b, c, and, d are nonnegative integers, is the
[#permalink]
Show Tags
26 Mar 2016, 17:57
I'm slightly confused. If a or d = 0, is ad / (10)^n = terminating decimal? Is 0 a terminating decimal?



Manager
Joined: 07 Jun 2018
Posts: 104

Given that a, b, c, and, d are nonnegative integers, is the
[#permalink]
Show Tags
19 Jul 2018, 13:56
Can someone please solve the second statement and show the working? b=(1+a)(a^2 2a + 1)  (a1)(a^2 1 ) Thanks!



Retired Moderator
Joined: 22 Aug 2013
Posts: 1416
Location: India

Re: Given that a, b, c, and, d are nonnegative integers, is the
[#permalink]
Show Tags
19 Jul 2018, 22:40
MrJglass wrote: Can someone please solve the second statement and show the working? b=(1+a)(a^2 2a + 1)  (a1)(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 (a1) and can thus be written as (a1)^2 or (a1)(a1). a^2  1 can be read as a^2  1^2 and can thus be written as (a+1)(a1) So now we can rewrite b as: b = (a+1)(a1)(a1)  (a1)(a+1)(a1) As you can see, both terms are same and thus their difference will be 0.



Manager
Joined: 07 Jun 2018
Posts: 104

Given that a, b, c, and, d are nonnegative integers, is the
[#permalink]
Show Tags
24 Jul 2018, 05:03
amanvermagmat wrote: MrJglass wrote: Can someone please solve the second statement and show the working? b=(1+a)(a^2 2a + 1)  (a1)(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 (a1) and can thus be written as (a1)^2 or (a1)(a1). a^2  1 can be read as a^2  1^2 and can thus be written as (a+1)(a1) So now we can rewrite b as: b = (a+1)(a1)(a1)  (a1)(a+1)(a1) 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!



NonHuman User
Joined: 09 Sep 2013
Posts: 13610

Re: Given that a, b, c, and, d are nonnegative integers, is the
[#permalink]
Show Tags
22 Oct 2019, 09:22
Hello from the GMAT Club BumpBot! Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up  doing my job. I think you may find it valuable (esp those replies with Kudos). Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________




Re: Given that a, b, c, and, d are nonnegative integers, is the
[#permalink]
22 Oct 2019, 09:22






