Author 
Message 
TAGS:

Hide Tags

Intern
Joined: 31 Oct 2010
Posts: 30

If 1=<n<=99, what is the probability that n(n + 1) is perfec
[#permalink]
Show Tags
11 Dec 2010, 23:52
Question Stats:
71% (00:49) correct 29% (01:31) wrong based on 53 sessions
HideShow timer Statistics
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?




Math Expert
Joined: 02 Sep 2009
Posts: 47978

Re: Beginner's Forum Question
[#permalink]
Show Tags
12 Dec 2010, 01:17
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.
_________________
New to the Math Forum? Please read this: Ultimate GMAT Quantitative Megathread  All You Need for Quant  PLEASE READ AND FOLLOW: 12 Rules for Posting!!! Resources: GMAT Math Book  Triangles  Polygons  Coordinate Geometry  Factorials  Circles  Number Theory  Remainders; 8. Overlapping Sets  PDF of Math Book; 10. Remainders  GMAT Prep Software Analysis  SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS)  Tricky questions from previous years.
Collection of Questions: PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
What are GMAT Club Tests? Extrahard Quant Tests with Brilliant Analytics




Manager
Joined: 11 Jul 2010
Posts: 208

Re: Beginner's Forum Question
[#permalink]
Show Tags
12 Dec 2010, 00:08
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: 30

Re: Beginner's Forum Question
[#permalink]
Show Tags
12 Dec 2010, 00:22
thanks for the quick reply! kudos for the help.



Senior Manager
Status: Bring the Rain
Joined: 17 Aug 2010
Posts: 358
Location: United States (MD)
Concentration: Strategy, Marketing
Schools: Michigan (Ross)  Class of 2014
GPA: 3.13
WE: Corporate Finance (Aerospace and Defense)

Re: Beginner's Forum Question
[#permalink]
Show Tags
13 Dec 2010, 08:41
Ah, that 30 sec approach looks good.
_________________
Go Blue!
GMAT Club Premium Membership  big benefits and savings



Manager
Joined: 14 Nov 2011
Posts: 131
Location: United States
Concentration: General Management, Entrepreneurship
GPA: 3.61
WE: Consulting (Manufacturing)

Re: Beginner's Forum Question
[#permalink]
Show Tags
07 Jul 2013, 04:12
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: 47978

Re: If 1=<n<=99, what is the probability that n(n + 1) is perfec
[#permalink]
Show Tags
07 Jul 2013, 06:36
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
_________________
New to the Math Forum? Please read this: Ultimate GMAT Quantitative Megathread  All You Need for Quant  PLEASE READ AND FOLLOW: 12 Rules for Posting!!! Resources: GMAT Math Book  Triangles  Polygons  Coordinate Geometry  Factorials  Circles  Number Theory  Remainders; 8. Overlapping Sets  PDF of Math Book; 10. Remainders  GMAT Prep Software Analysis  SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS)  Tricky questions from previous years.
Collection of Questions: PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
What are GMAT Club Tests? Extrahard Quant Tests with Brilliant Analytics



Current Student
Joined: 04 Jul 2014
Posts: 302
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, 03:19
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
_________________
Cheers!!
JA If you like my post, let me know. Give me a kudos!



Target Test Prep Representative
Status: Founder & CEO
Affiliations: Target Test Prep
Joined: 14 Oct 2015
Posts: 3180
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, 18:00
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
_________________
Scott WoodburyStewart
Founder and CEO
GMAT Quant SelfStudy Course
500+ lessons 3000+ practice problems 800+ HD solutions



Manager
Joined: 20 Jan 2017
Posts: 60
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, 20:05
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: 7749

Re: If 1=<n<=99, what is the probability that n(n + 1) is perfec
[#permalink]
Show Tags
06 Apr 2018, 20:04
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.
_________________
GMAT Books  GMAT Club Tests  Best Prices on GMAT Courses  GMAT Mobile App  Math Resources  Verbal Resources




Re: If 1=<n<=99, what is the probability that n(n + 1) is perfec &nbs
[#permalink]
06 Apr 2018, 20:04






