Author 
Message 
TAGS:

Hide Tags

Intern
Joined: 31 Oct 2010
Posts: 28

If 1=<n<=99, what is the probability that n(n + 1) is perfec
[#permalink]
Show Tags
11 Dec 2010, 22:52
Question Stats:
45% (01:33) correct 55% (01:40) wrong based on 122 sessions
HideShow timer Statistics
If \(1 \leq n \leq 99\), what is the probability that n(n + 1) is divisible by 3 ? A. 1/4 B. 1/3 C. 1/2 D. 2/3 E. 3/4 For n(n + 1) to be divisible by 3, either n is a multiple of 3, or (n + 1) is a multiple of 3.
Considering n is a multiple of 3: Between 1 to 99, the number of multiples of 3 = (99  3)/3 + 1 = 33 [Note: Use I substraced 3 from 99 since the smallest multiple of 3 is 3 itself]
Considering (n + 1) is a multiple of 3: Between 1 to 99, the number of multiples of 3 would be from 298 and would also be 33 multiples.
Thus, the probability that n(n+1) is perfectly divisible by 3 = P(n is odd) or P(n + 1 is odd) = 33/99 + 33/99 = 66/99 = 22/33 = 2/3
I am not sure of the underlying formula used here. Am I right assuming to find the number of multiples divisble (nL(last)nF(first))/3(multiple of) +1 and I add 1 because i need to account for 3 again?
Official Answer and Stats are available only to registered users. Register/ Login.




Math Expert
Joined: 02 Sep 2009
Posts: 64092

Re: If 1=<n<=99, what is the probability that n(n + 1) is perfec
[#permalink]
Show Tags
12 Dec 2010, 00:17
mmcooley33 wrote: If 1=<n<=99, what is the probability that n(n + 1) is perfectly divisible by 3 ?
For n(n + 1) to be divisible by 3, either n is a multiple of 3, or (n + 1) is a multiple of 3.
Considering n is a multiple of 3: Between 1 to 99, the number of multiples of 3 = (99  3)/3 + 1 = 33 [Note: Use I substraced 3 from 99 since the smallest multiple of 3 is 3 itself]
Considering (n + 1) is a multiple of 3: Between 1 to 99, the number of multiples of 3 would be from 298 and would also be 33 multiples.
Thus, the probability that n(n+1) is perfectly divisible by 3 = P(n is odd) or P(n + 1 is odd) = 33/99 + 33/99 = 66/99 = 22/33 = 2/3
I am not sure of the underlying formula used here. Am I right assuming to find the number of multiples divisble (nL(last)nF(first))/3(multiple of) +1 and I add 1 because i need to account for 3 again? First about the multiple of x in the given range:\(# \ of \ multiples \ of \ x \ in \ the \ range = \frac{Last \ multiple \ of \ x \ in \ the \ range \  \ First \ multiple \ of \ x \ in \ the \ range}{x}+1\), (check this: totallybasic94862.html?hilit=multiple%20range). So, there will be \(\frac{993}{3}+1=33\) multiples of 3 in the range from 1 to 99, inclusive: 3, 6, 9, ..., 99; n+1 to be a multiple of 3 in the range from 1 to 99, inclusive n must be 2, 5, 8, ..., 98, one less than above values, so also 33 options for n; Hence, n(n+1) to be a multiple of 3 n can take 33+33=66 values: P=(favorable outcomes)/(total # of outcomes)=66/99=2/3. 30 second approach:Take first 3 numbers from the given range: 1, 2, and 3 in order n(n+1) to be a multiple of 3 n can be either 2 or 3 P=2/3. For next 3 numbers the probability will be the same and as 99 equals to 3*33 (similar 33 blocks of 3) then for complete range the probability will still be 2/3. Answer: 2/3. Hope it helps.
_________________




Manager
Joined: 11 Jul 2010
Posts: 150

Re: If 1=<n<=99, what is the probability that n(n + 1) is perfec
[#permalink]
Show Tags
11 Dec 2010, 23:08
this formula is in the MGMAT math book.. really useful to know... you add 1 as you generally underestimate by 1 at the extreme  this evenly spaced sets formula is the basis for the simpler formula: (lastFirst) + 1 when you have to count out the number of elements in a consecutive sequence 1, 2, 3.... You are basically doing (lastfirst)/1 + 1 > you are basically dividing by 1 there as that is the common space b/w the elements... Knowing that you can apply this formula for evenly spaced sets should be enough...



Intern
Joined: 31 Oct 2010
Posts: 28

Re: If 1=<n<=99, what is the probability that n(n + 1) is perfec
[#permalink]
Show Tags
11 Dec 2010, 23:22
thanks for the quick reply! kudos for the help.



Senior Manager
Status: Bring the Rain
Joined: 17 Aug 2010
Posts: 333
Location: United States (MD)
Concentration: Strategy, Marketing
Schools: Michigan (Ross)  Class of 2014
GPA: 3.13
WE: Corporate Finance (Aerospace and Defense)

Re: If 1=<n<=99, what is the probability that n(n + 1) is perfec
[#permalink]
Show Tags
13 Dec 2010, 07:41
Ah, that 30 sec approach looks good.



Manager
Joined: 14 Nov 2011
Posts: 113
Location: United States
Concentration: General Management, Entrepreneurship
GPA: 3.61
WE: Consulting (Manufacturing)

Re: If 1=<n<=99, what is the probability that n(n + 1) is perfec
[#permalink]
Show Tags
07 Jul 2013, 03:12
Bunuel wrote: mmcooley33 wrote: If 1=<n<=99, what is the probability that n(n + 1) is perfectly divisible by 3 ?
For n(n + 1) to be divisible by 3, either n is a multiple of 3, or (n + 1) is a multiple of 3.
Considering n is a multiple of 3: Between 1 to 99, the number of multiples of 3 = (99  3)/3 + 1 = 33 [Note: Use I substraced 3 from 99 since the smallest multiple of 3 is 3 itself]
Considering (n + 1) is a multiple of 3: Between 1 to 99, the number of multiples of 3 would be from 298 and would also be 33 multiples.
Thus, the probability that n(n+1) is perfectly divisible by 3 = P(n is odd) or P(n + 1 is odd) = 33/99 + 33/99 = 66/99 = 22/33 = 2/3
I am not sure of the underlying formula used here. Am I right assuming to find the number of multiples divisble (nL(last)nF(first))/3(multiple of) +1 and I add 1 because i need to account for 3 again? First about the multiple of x in the given range:\(# \ of \ multiples \ of \ x \ in \ the \ range = \frac{Last \ multiple \ of \ x \ in \ the \ range \  \ First \ multiple \ of \ x \ in \ the \ range}{x}+1\), (check this: totallybasic94862.html?hilit=multiple%20range). So, there will be \(\frac{993}{3}+1=33\) multiples of 3 in the range from 1 to 99, inclusive: 3, 6, 9, ..., 99; n+1 to be a multiple of 3 in the range from 1 to 99, inclusive n must be 2, 5, 8, ..., 98, one less than above values, so also 33 options for n; Hence, n(n+1) to be a multiple of 3 n can take 33+33=66 values: P=(favorable outcomes)/(total # of outcomes)=66/99=2/3. 30 second approach:Take first 3 numbers from the given range: 1, 2, and 3 in order n(n+1) to be a multiple of 3 n can be either 2 or 3 P=2/3. For next 3 numbers the probability will be the same and as 99 equals to 3*33 (similar 33 blocks of 3) then for complete range the probability will still be 2/3. Answer: 2/3. Hope it helps. Bump for the 30 sec approach.



Math Expert
Joined: 02 Sep 2009
Posts: 64092

Re: If 1=<n<=99, what is the probability that n(n + 1) is perfec
[#permalink]
Show Tags
07 Jul 2013, 05:36
mmcooley33 wrote: If 1 <n <99, what is the probability that n(n + 1) is perfectly divisible by 3 ? For n(n + 1) to be divisible by 3, either n is a multiple of 3, or (n + 1) is a multiple of 3.
Considering n is a multiple of 3: Between 1 to 99, the number of multiples of 3 = (99  3)/3 + 1 = 33 [Note: Use I substraced 3 from 99 since the smallest multiple of 3 is 3 itself]
Considering (n + 1) is a multiple of 3: Between 1 to 99, the number of multiples of 3 would be from 298 and would also be 33 multiples.
Thus, the probability that n(n+1) is perfectly divisible by 3 = P(n is odd) or P(n + 1 is odd) = 33/99 + 33/99 = 66/99 = 22/33 = 2/3
I am not sure of the underlying formula used here. Am I right assuming to find the number of multiples divisble (nL(last)nF(first))/3(multiple of) +1 and I add 1 because i need to account for 3 again? Similar questions to practice: ifanintegernistobechosenatrandomfromtheintegers126654.htmlifintegercisrandomlyselectedfrom20to99inclusive121561.html
_________________



Senior Manager
Joined: 04 Jul 2014
Posts: 293
Location: India
GMAT 1: 640 Q47 V31 GMAT 2: 640 Q44 V34 GMAT 3: 710 Q49 V37
GPA: 3.58
WE: Analyst (Accounting)

Re: If 1=<n<=99, what is the probability that n(n + 1) is perfec
[#permalink]
Show Tags
22 Nov 2014, 02:19
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 There's a discussion going on here for this question: anintegernbetween1and99inclusiveistobechosenat160998.html#p1445924



Target Test Prep Representative
Status: Founder & CEO
Affiliations: Target Test Prep
Joined: 14 Oct 2015
Posts: 10542
Location: United States (CA)

Re: If 1=<n<=99, what is the probability that n(n + 1) is perfec
[#permalink]
Show Tags
16 Jan 2017, 17:00
mmcooley33 wrote: If 1<n<99, what is the probability that n(n + 1) is perfectly divisible by 3 ? We are given that an integer n is to be selected at random from 1 to 99 inclusive, and we must determine the probability that n(n+1) will be divisible by 3. Since probability = favorable outcomes/total outcomes and we know that the total number of outcomes is 99, because there are 99 integers from 1 to 99 inclusive, we need to determine the number of values of n such that n(n+1) is divisible by 3. First, we can determine the number of values of n that are divisible by 3, that is, the number of multiples of 3 that are between 3 and 99 inclusive. To calculate this, we can use the formula: (Largest multiple of 3 – smallest multiple of 3)/3 + 1 (99  3)/3 + 1 96/3 + 1 32 + 1 = 33 Thus, there are 33 multiples of 3 between 1 and 99 inclusive. That is, the value of n can be any one of these 33 multiples of 3 so that n(n + 1) will be divisible by 3. Similarly, if n + 1 is a multiple of 3, n(n + 1) will be also be divisible by 3. Since we know that there are 33 values of n that are multiples of 3, there must be another 33 values of n such that n + 1 is a multiple of 3. Let’s expand on this idea: When n = 2, n + 1 = 3, and thus n(n+1) is a multiple of 3. When n = 23, n + 1 = 24, and thus n(n+1) is a multiple of 3. When n = 98, n + 1 = 99, and thus n(n+1) is a multiple of 3. We can see that there are 33 values of n that are multiples of 3, and 33 more values of n for n + 1 to be multiples of 3. Thus, the probability of selecting a value of n so that n(n+1) is a multiple of 3 is: 66/99 = 2/3
_________________
5star rated online GMAT quant self study course See why Target Test Prep is the top rated GMAT quant course on GMAT Club. Read Our Reviews If you find one of my posts helpful, please take a moment to click on the "Kudos" button.



Current Student
Joined: 20 Jan 2017
Posts: 49
Location: United States (NY)
GMAT 1: 750 Q48 V44 GMAT 2: 610 Q34 V41
GPA: 3.92

Re: If 1=<n<=99, what is the probability that n(n + 1) is perfec
[#permalink]
Show Tags
24 Jan 2017, 19:05
1) n(n+1) is divisible by 3 when either n is divisible by 3 or (n+1) is divisible by 3. 2) n is divisible by 3 in 99/3=33 cases 3) n+1 is divisible by 3 in 99/3=33 cases 4) 33+33=66 5) 66/99=2/3



NonHuman User
Joined: 09 Sep 2013
Posts: 14969

Re: If 1=<n<=99, what is the probability that n(n + 1) is
[#permalink]
Show Tags
31 Jul 2019, 00:45
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: If 1=<n<=99, what is the probability that n(n + 1) is
[#permalink]
31 Jul 2019, 00:45




