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# An integer n between 1 and 99, inclusive, is to be chosen at

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

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04 Oct 2013, 03:12
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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
[Reveal] Spoiler: OA

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Last edited by Bunuel on 07 Nov 2013, 09:54, edited 2 times in total.
Renamed the topic, edited the question and added the OA.

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

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04 Oct 2013, 04:34
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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.

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

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26 Dec 2013, 23:54
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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
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Re: An integer n between 1 and 99, inclusive, is to be chosen at [#permalink]

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12 Apr 2014, 23:56
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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

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

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17 Aug 2014, 00:19
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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

Last edited by Raihanuddin on 22 Feb 2015, 09:52, edited 1 time in total.

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

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10 Sep 2014, 00:54
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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;

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

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22 Nov 2014, 03:16
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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
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Re: An integer n between 1 and 99, inclusive, is to be chosen at [#permalink]

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13 Feb 2015, 06:51
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.

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

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13 Feb 2015, 08:33
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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.

Last edited by Raihanuddin on 21 Jul 2015, 07:00, edited 1 time in total.

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

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13 Feb 2015, 09:26
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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 !!

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

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13 Feb 2015, 09:37
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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.

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

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13 Feb 2015, 09:59
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!

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

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16 Feb 2015, 03:29
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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
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An integer n between 1 and 99, inclusive, is to be chosen at [#permalink]

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05 May 2015, 09:59
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

Last edited by AjChakravarthy on 05 May 2015, 20:30, edited 1 time in total.

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

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05 May 2015, 12:00
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

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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, 13:06
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.

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

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27 May 2015, 21:30
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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...

[Reveal] Spoiler:
D

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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, 22:06
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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, 18:42
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, 10:52
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

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An integer n between 1 and 99, inclusive, is to be chosen at   [#permalink] 23 Aug 2017, 10:52
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