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• ### $450 Tuition Credit & Official CAT Packs FREE November 15, 2018 November 15, 2018 10:00 PM MST 11:00 PM MST EMPOWERgmat is giving away the complete Official GMAT Exam Pack collection worth$100 with the 3 Month Pack ($299) # An integer n between 1 and 99, inclusive, is to be chosen at  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics Author Message TAGS: ### Hide Tags Intern Joined: 29 May 2012 Posts: 9 Concentration: General Management, Finance GPA: 3.49 An integer n between 1 and 99, inclusive, is to be chosen at [#permalink] ### Show Tags Updated on: 07 Nov 2013, 08:54 9 1 89 00:00 Difficulty: 45% (medium) Question Stats: 67% (01:01) correct 33% (01:17) wrong based on 1487 sessions ### HideShow timer Statistics An integer n between 1 and 99, inclusive, is to be chosen at random. What is the probability that n(n+1) will be divisible by 3 ? A) 1/9 B) 1/3 C) 1/2 D) 2/3 E) 5/6 _________________ "When ideas get really complicated and the world gets complicated, its foolish to think that the person who's first can figure it all out" Originally posted by accincognito on 04 Oct 2013, 02:12. Last edited by Bunuel on 07 Nov 2013, 08:54, edited 2 times in total. Renamed the topic, edited the question and added the OA. ##### Most Helpful Expert Reply Math Expert Joined: 02 Sep 2009 Posts: 50544 Re: An integer n between 1 and 99, inclusive, is to be chosen at [#permalink] ### Show Tags 04 Oct 2013, 03:34 40 41 paskorntt wrote: An integer n between 1 and 99, inclusive, is to be chosen at random. What is the probability that n(n+1) will be divisible by 3 ? A) 1/9 B) 1/3 C) 1/2 D) 2/3 E) 5/6 n(n+1) to be divisible by 3 either n or n+1 must be a multiples of 3. In each following group of numbers: {1, 2, 3}, {4, 5, 6}, {7, 8, 9}, ..., {97, 98, 99} there are EXACTLY 2 numbers out of 3 satisfying the above condition. For example in {1, 2, 3} n can be: 2, or 3. Thus, the overall probability is 2/3. Answer: D. _________________ ##### Most Helpful Community Reply Manager Joined: 11 Sep 2013 Posts: 157 Concentration: Finance, Finance An integer n between 1 and 99, inclusive, is to be chosen at [#permalink] ### Show Tags Updated on: 22 Feb 2015, 08:52 23 23 In such problem always follow a simple rule. The divisor is 3. So take number from 1 to 3 as example If n=1, n(n+1)= Not divisible by 3 If n=2, n(n+1)= yes If n=3, n(n+1)=yes. Don't need any more example. So, 2 out of 3 are divisible. Hence, 2/3 is the answer. This rule is applicable if last number(99 in this case) is divisible by 3. If you had to pick from 1-98, still you can apply the rule but be careful. There is one more additional step. Hopefully you can find answer now in less than 30 seconds Originally posted by Raihanuddin on 16 Aug 2014, 23:19. Last edited by Raihanuddin on 22 Feb 2015, 08:52, edited 1 time in total. ##### General Discussion Senior Manager Joined: 23 Oct 2010 Posts: 353 Location: Azerbaijan Concentration: Finance Schools: HEC '15 (A) GMAT 1: 690 Q47 V38 Re: An integer n between 1 and 99, inclusive, is to be chosen at [#permalink] ### Show Tags 26 Dec 2013, 22:54 6 2 the required number = (if n is even )+ (if n is divisible by 3) - (if n is even and divisible by 3)= =(49+33-16)/99=66/99=2/3 _________________ Happy are those who dream dreams and are ready to pay the price to make them come true I am still on all gmat forums. msg me if you want to ask me smth Intern Joined: 29 Mar 2014 Posts: 12 Location: United States Concentration: Entrepreneurship, Finance GMAT 1: 720 Q50 V39 GPA: 3 Re: An integer n between 1 and 99, inclusive, is to be chosen at [#permalink] ### Show Tags 12 Apr 2014, 22:56 1 3 Any three consecutive integers will exactly have one number that is a multiple of three. So the probability of 2 consecutive integers to have one multiple of 3 is 2/3 (if the total set of numbers considered is a multiple of three, it doesn't matter if the set is {1,2..6}, {1,2,..30}, or {1,2,3...99}, the probability will always be 2/3). Ans: D Manager Status: suffer now and live forever as a champion!!! Joined: 01 Sep 2013 Posts: 114 Location: India Dheeraj: Madaraboina GPA: 3.5 WE: Information Technology (Computer Software) Re: An integer n between 1 and 99, inclusive, is to be chosen at [#permalink] ### Show Tags 09 Sep 2014, 23:54 2 Hi guys, Could anyone tell whether my approach for this problem is correct. Multiples of 3 between 1 and 99 are 33. Multiples of (n+1) between 1 and 99 must also be 33. Probability that [n(n+1)/3 ] is divisible by 3 is either n is divisible by 3 (or) (n+1) is divisible by 3 . Probablity that n is divisible by 3 is 33/99 = 1/3; Probabilty that n+1 is divisible by 3 is 33/99 = 1/3; Therefore the probability that n(n+1)/3 is 1/3+ 1/3 = 2/3; Thanks in advance Current Student Joined: 04 Jul 2014 Posts: 298 Location: India GMAT 1: 640 Q47 V31 GMAT 2: 640 Q44 V34 GMAT 3: 710 Q49 V37 GPA: 3.58 WE: Analyst (Accounting) Re: An integer n between 1 and 99, inclusive, is to be chosen at [#permalink] ### Show Tags 22 Nov 2014, 02:16 3 1 accincognito wrote: An integer n between 1 and 99, inclusive, is to be chosen at random. What is the probability that n(n+1) will be divisible by 3 ? A) 1/9 B) 1/3 C) 1/2 D) 2/3 E) 5/6 Wow! I did this one very fast and correctly! :D (After getting a series of other questions in this type wrong) Here is a very simple solution. For n(n+1) to be divisible by 3, either a) n must be a multiple of 3 --> Number of possibilities = 99/3 = 33 OR b) (n -1) must be a multiple of 3 --> Number of possibilities = 99/3 = 33 Now, favorable possibilities = 66 and total possibilities = 100 (100 because 99 and 1 are both included). Therefore, probability = 66/100 or 2/3 :D _________________ Cheers!! JA If you like my post, let me know. Give me a kudos! Intern Joined: 31 Oct 2013 Posts: 12 Schools: INSEAD Jan '16 Re: An integer n between 1 and 99, inclusive, is to be chosen at [#permalink] ### Show Tags 13 Feb 2015, 05:51 Hi Bunue/ Raihanuddin Can you use the same theory for n(n+1) divisible by 8. For example take (1,2,3,4,5,6,7,8), only 7 and 8 work so prob is 2/8 = 1/4? Also how would you do it or even n(n+1)(n-1) divisible by 8 ? Is there a general rule/trick, might help a lot to reduce time spent on these questions. Manager Joined: 11 Sep 2013 Posts: 157 Concentration: Finance, Finance An integer n between 1 and 99, inclusive, is to be chosen at [#permalink] ### Show Tags Updated on: 21 Jul 2015, 06:00 10 13 Onthemove wrote: Hi Bunue/ Raihanuddin Can you use the same theory for n(n+1) divisible by 8. For example take (1,2,3,4,5,6,7,8), only 7 and 8 work so prob is 2/8 = 1/4? Also how would you do it or even n(n+1)(n-1) divisible by 8 ? Is there a general rule/trick, might help a lot to reduce time spent on these questions. If it is by 8, then you have to do the above plus an additional step. if n = 97, n(n+1) not divisible by 8 if n= 98 still not divisible by 8 and if n= 99 also not divisible by 8 Now notice that from 1 to 96, total 24 values of n is divisible by 8. We have 3 more values as shown above. So out of 99 values 24 are divisible. Probability = 24/99 = 8/33 You may think that where I have got 96. from 1 to 99 we can't divide 99 by 8. The highest number we can divide is 96. So we have three more numbers left. That's why we have to do the additional step. We are using cycle of (1-8), (9-16) (17-24) .......... You can see that 2 numbers from (1-8) are divisible by 8 if you consider n(n+1). Similarly 2 numbers from (9-16) and so on How many such pairs do we have from 1-99.? 12 pairs of 8. So 2 numbers from each pair are divisible. Therefore total 2*12= 24 are divisible out of 96 numbers. And from remaining number none is divisible by 8. So, out of 99 numbers 24 are divisible. You can do the next question similarly. Very important I think in official question you will not get such number as 8 by which you can't divide last number, which is 99. So, don't worry for that. The given question in the main post is an official question and I have always got that GMAT gives you last number divisible by the divisor. Originally posted by Raihanuddin on 13 Feb 2015, 07:33. Last edited by Raihanuddin on 21 Jul 2015, 06:00, edited 1 time in total. Intern Joined: 31 Oct 2013 Posts: 12 Schools: INSEAD Jan '16 Re: An integer n between 1 and 99, inclusive, is to be chosen at [#permalink] ### Show Tags 13 Feb 2015, 08:26 1 hi, Thanks for replying so quickly. yes the above question is official. I did the GMAT Exam Pack PT 3 today, this was one of the questions. I understand your method completely. If the last number in the sequence(in our case 99), is not divisible by 8 then I will use the pairing method, its easy enough. 2/8 = x/96, in this case. Its just a simple ratio of pairs. And if it works out as in most official questions then even better. Either way i can comfortably do the problem in a 90 seconds at the most. So great solution mate !! Manager Joined: 11 Sep 2013 Posts: 157 Concentration: Finance, Finance Re: An integer n between 1 and 99, inclusive, is to be chosen at [#permalink] ### Show Tags 13 Feb 2015, 08:37 1 I am happy that the method has worked for you. I got 50 in Quant. Hope to get 51 in future. But at this moment I am not at all concerned about GMAT. I think if you just follow official questions(all available prep in GC +exam pack+question pack) and understand the concept well, you are sure to get at least 50. To build your concept you can do 500 and 600 level question from GC. Bunnuel is awesome. Intern Joined: 31 Oct 2013 Posts: 12 Schools: INSEAD Jan '16 Re: An integer n between 1 and 99, inclusive, is to be chosen at [#permalink] ### Show Tags 13 Feb 2015, 08:59 Yes agreed all the solutions on GMAT CLUB are tremendously helpful. I check most of Bunuel's solutions for tough problems. Makes it seem so easy though, LOL! Math Expert Joined: 02 Sep 2009 Posts: 50544 Re: An integer n between 1 and 99, inclusive, is to be chosen at [#permalink] ### Show Tags 16 Feb 2015, 02:29 3 7 accincognito wrote: An integer n between 1 and 99, inclusive, is to be chosen at random. What is the probability that n(n+1) will be divisible by 3 ? A) 1/9 B) 1/3 C) 1/2 D) 2/3 E) 5/6 Similar questions mimght help: if-integer-c-is-randomly-selected-from-20-to-99-inclusive-121561.html if-an-integer-n-is-to-be-chosen-at-random-from-the-integers-126654.html if-1-n-99-what-is-the-probability-that-n-n-1-is-perfec-106168.html if-integer-c-is-randomly-selected-from-20-to-99-inclusive-121561.html _________________ Manager Joined: 08 Oct 2013 Posts: 50 An integer n between 1 and 99, inclusive, is to be chosen at [#permalink] ### Show Tags Updated on: 05 May 2015, 19:30 So here either n or (n+1) can be a multiple of 3 1. No. of ways n can be a multiple of 3 is 33 2. No. of ways (n+1) can be a multiple of 3 is 33. (this is the tricky part; this is how I deduced it: For every multiple of 3, there exists a number that is 1 less than a multiple of 3. Therefore there are 33 numbers such that (n+1) is a multiple of 3) Total ways of desired/ Total ways possible = (33+33)/ 99 = 66/99 = 2/3 However i don't think this is a 55% difficulty level question. Must be way more than that. Regards, Aj Originally posted by AjChakravarthy on 05 May 2015, 08:59. Last edited by AjChakravarthy on 05 May 2015, 19:30, edited 1 time in total. Manager Joined: 05 May 2015 Posts: 73 Re: An integer n between 1 and 99, inclusive, is to be chosen at [#permalink] ### Show Tags 05 May 2015, 11:00 for n(n+1) to be divisible by 3, either n or n+1 is divisible by 3 33 numbers from 1 to 99 is divisible by 3 33 numbers from 1 to 99 will make n+1 divisible by 3 so (33+ 33)/ 99 = 2/3 Manager Joined: 23 Nov 2014 Posts: 56 Location: India GMAT 1: 730 Q49 V40 GPA: 3.14 WE: Sales (Consumer Products) An integer n between 1 and 99, inclusive, is to be chosen at [#permalink] ### Show Tags 27 May 2015, 12:06 Would appreciate some input regarding my confusion with this problem. If n(n+1) has to be divisible by 3, there could be 3 scenarios in all: 1) n is divisible by 3 AND (n+1) is not divisibly by 3 I see the mistake here now: if n is divisible by 3, n+1 is automatically not divisible by 3 2) n is not divisible by 3 AND (n+1) is divisible by 3 Same error as above 3) n AND (n+1) are BOTH divisible by 3 We're asked to pick only one integer n so this is invalid/unnecessary Out of 99 elements, 33 will be of the form n divisible by 3, 33 will be of the form (n+1) divisibly by 3, and 33 will be neither. So the total probability for cases 1-3 above, per me, should work out to something like: (33/99*33/99)*3 = (1/9)*3 = 1/3. Please tell me what part of the above logic is flawed? Thank you. Edit: Have retraced my steps and commented what I think my mistakes were - would still appreciate some endorsement from other members. EMPOWERgmat Instructor Status: GMAT Assassin/Co-Founder Affiliations: EMPOWERgmat Joined: 19 Dec 2014 Posts: 12841 Location: United States (CA) GMAT 1: 800 Q51 V49 GRE 1: Q170 V170 Re: An integer n between 1 and 99, inclusive, is to be chosen at [#permalink] ### Show Tags 27 May 2015, 20:30 4 1 Hi All, This question is based on pattern-matching. The good news is that even if you don't immediately spot the pattern (some people can just "see it"), you can still PROVE that a pattern exists with just a little work... We're told that N is a number from 1 to 99, inclusive. We're asked for the probability that N(N+1) is divisible by 3.... IF.... N = 1....(1)(2) = 2 is NOT divisible by 3. N = 2....(2)(3) = 6 IS divisible by 3 N = 3....(3)(4) = 12 IS divisible by 3. In these first 3 examples, notice how 2 of the 3 are divisible by 3....What happens when we look at the NEXT 3 numbers.... IF.... N = 4....(4)(5) = 20 is NOT divisible by 3. N = 5....(5)(6) = 30 IS divisible by 3 N = 6....(6)(7) = 42 IS divisible by 3. The pattern REPEATS! This means that for ever 3 consecutive values of N, 2 of the 3 will be divisible by 3. Since we're dealing with the numbers 1 to 99, inclusive, that means we'll have 33 "sets" of 3 numbers (as shown above). Thus, 2/3 of ALL the values WILL be divisible by 3... Final Answer: GMAT assassins aren't born, they're made, Rich _________________ 760+: Learn What GMAT Assassins Do to Score at the Highest Levels Contact Rich at: Rich.C@empowergmat.com # Rich Cohen Co-Founder & GMAT Assassin Special Offer: Save$75 + GMAT Club Tests Free
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An integer n between 1 and 99, inclusive, is to be chosen at  [#permalink]

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27 May 2015, 21:06
2
1
avgroh wrote:
Would appreciate some input regarding my confusion with this problem.

If n(n+1) has to be divisible by 3, there could be 3 scenarios in all:

1) n is divisible by 3 AND (n+1) is not divisibly by 3 I see the mistake here now: if n is divisible by 3, n+1 is automatically not divisible by 3

2) n is not divisible by 3 AND (n+1) is divisible by 3 Same error as above

3) n AND (n+1) are BOTH divisible by 3 We're asked to pick only one integer n so this is invalid/unnecessary

Out of 99 elements, 33 will be of the form n divisible by 3, 33 will be of the form (n+1) divisibly by 3, and 33 will be neither.

So the total probability for cases 1-3 above, per me, should work out to something like: (33/99*33/99)*3 = (1/9)*3 = 1/3.

Please tell me what part of the above logic is flawed?

Thank you.

Edit: Have retraced my steps and commented what I think my mistakes were - would still appreciate some endorsement from other members.

Hi avgroh,

You have rightly calculated that there will be 33 numbers of the form n which are divisible by 3 and 33 numbers of the form n + 1 which are divisible by 3.

The question asks us the probability of product of two numbers being divisible by 3. This is possible if either of them is divisible by 3 i.e. if n is divisible by 3 or n + 1 is divisible by 3. So, it's an OR event. We don't need both the events to happen for n(n+1) to be divisible by 3. Hence we will add the probabilities of both the events happening.

So we can write P(n(n+1)) is divisible by 3 $$= \frac{33}{99} + \frac{33}{99} = \frac{66}{99} = \frac{2}{3}$$

Hope it's clear

Regards
Harsh
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Re: An integer n between 1 and 99, inclusive, is to be chosen at  [#permalink]

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22 Jul 2017, 17:42
accincognito wrote:
An integer n between 1 and 99, inclusive, is to be chosen at random. What is the probability that n(n+1) will be divisible by 3 ?

A) 1/9
B) 1/3
C) 1/2
D) 2/3
E) 5/6

For any three consecutive integers, the probability that selecting a number and its consecutive integer will produce a multiple of 3 is 2/3.

Consider the following case:

1,2,3

If n = 1, then n+1 = 2. 2*1 => Not divisible by three.
If n = 2, then n+1 = 3 2*3 => Divisible by three.
If n = 3, then n+1 = 4 3*4 => Divisible by three.

2/3
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An integer n between 1 and 99, inclusive, is to be chosen at  [#permalink]

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23 Aug 2017, 09:52
Hi Bunuel
I solved the sum using the following approach, but got it wrong. Can you please explain the error in my approach.
For n(n+1) to be divisible by 3, either
a) n must be a multiple of 3 --> Number of possibilities = 96/3 = 32 (because if we consider n=99 then n+1 will not lie between 1 and 99, so the last number that n can be is 96)OR
b) (n +1) must be a multiple of 3 --> Number of possibilities = 99/3 = 33

Now, favorable possibilities = 65 and total possibilities = 99

Therefore, probability = 65/99
An integer n between 1 and 99, inclusive, is to be chosen at &nbs [#permalink] 23 Aug 2017, 09:52

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