Oct 19 07:00 AM PDT  09:00 AM PDT Does GMAT RC seem like an uphill battle? eGMAT is conducting a free webinar to help you learn reading strategies that can enable you to solve 700+ level RC questions with at least 90% accuracy in less than 10 days. Sat., Oct 19th at 7 am PDT Oct 18 08:00 AM PDT  09:00 AM PDT Learn an intuitive, systematic approach that will maximize your success on Fillintheblank GMAT CR Questions. Oct 20 07:00 AM PDT  09:00 AM PDT Get personalized insights on how to achieve your Target Quant Score. Oct 22 08:00 PM PDT  09:00 PM PDT On Demand for $79. For a score of 4951 (from current actual score of 40+) AllInOne Standard & 700+ Level Questions (150 questions) Oct 23 08:00 AM PDT  09:00 AM PDT Join an exclusive interview with the people behind the test. If you're taking the GMAT, this is a webinar you cannot afford to miss!
Author 
Message 
TAGS:

Hide Tags

Intern
Joined: 07 Nov 2012
Posts: 12

If each participant of a chess tournament plays exactly one
[#permalink]
Show Tags
10 Nov 2012, 14:06
Question Stats:
71% (01:49) correct 29% (02:05) wrong based on 370 sessions
HideShow timer Statistics
If each participant of a chess tournament plays exactly one game with each of the remaining participants, then 153 games will be played during the tournament. Find the number of participants. A. 15 B. 16 C. 17 D. 18 E. 19
Official Answer and Stats are available only to registered users. Register/ Login.




Math Expert
Joined: 02 Sep 2009
Posts: 58453

Re: If each participant of a chess tournament plays exactly one
[#permalink]
Show Tags
11 Nov 2012, 05:40
derekgmat wrote: If each participant of a chess tournament plays exactly one game with each of the remaining participants, then 153 games will be played during the tournament. Find the number of participants.
A. 15 B. 16 C. 17 D. 18 E. 19 We are basically told that we can choose 153 groups of two players out of n players, thus \(C^2_n=153\) > \(\frac{n!}{(n2)!2!}=153\) > \(\frac{(n1)n}{2}=153\) > \((n1)n=306\) > \(n=18\). Answer: D. Similar questions to practice: howmanydiagonalsdoesapolygonwith21sideshaveifone101540.htmlif10personsmeetatareunionandeachpersonshakeshands110622.html10businessexecutivesand7chairmenmeetataconference126163.htmlhowmanydifferenthandshakesarepossibleifsixgirls129992.html15chessplayerstakepartinatournamenteveryplayer55939.htmlthereare5chessamateursplayinginvillaschessclub127235.htmlHope it helps.
_________________




Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 9706
Location: Pune, India

Re: If each participant of a chess tournament plays exactly one
[#permalink]
Show Tags
10 Nov 2012, 19:00
derekgmat wrote: If each participant of a chess tournament plays exactly one game with each of the remaining participants, then 153 games will be played during the tournament. Find the number of participants.
A. 15 B. 16 C. 17 D. 18 E. 19 If the number of participants is 3 (say A, B, C) the number of games played will be 2 (A plays against B and C) + 1 (B plays against C) = 3 Using the same logic, if the number of participants is n, the number of games played will be (n1) + (n  2) + (n  3) + ... 3 + 2 + 1 Given that this sum = 153 = 1 + 2 + 3 + ... ( n  1) Sum of first m positive integers is given by m(m+1)/2. So sum of first (n1) positive integers is (n1)*n/2 153 = (n1)*n/2 (n1)*n = 306 17*18 = 306 (We know that 15^2 = 225 so the two consecutive numbers must be greater than 15. Also, 20^2 = 400 so the two numbers must be less than 20. The pair of numbers in between 15 and 20 whose product ends with 6 is 17 and 18) So n = 18 Answer (D)
_________________
Karishma Veritas Prep GMAT Instructor
Learn more about how Veritas Prep can help you achieve a great GMAT score by checking out their GMAT Prep Options >



Intern
Joined: 07 Nov 2012
Posts: 12

Re: If each participant of a chess tournament plays exactly one
[#permalink]
Show Tags
11 Nov 2012, 01:29
Thanks Karishma
I must admit I am struggling to work through your solution. I looked at the question again and answered it with the approach of working through each of the choices longhand.
If A, then 15 players means they each will play 14 games (so 14*15), divide the result by two for double counted games: 14*15 = 210 201/2 = 105, not 153, next
If B, 15*16 = 240 240/2 = 120, not 153, next
If C, 16*17 = 274 274/2 = 137, not 153, next
If D, 17*18 = 306 306/2 = 153 ANSWER D
Am I missing any important concepts by answering the question with this longhand method as opposed to your more structured approach?
Kind regards
Derek



Director
Status: Done with formalities.. and back..
Joined: 15 Sep 2012
Posts: 563
Location: India
Concentration: Strategy, General Management
Schools: Olin  Wash U  Class of 2015
WE: Information Technology (Computer Software)

Re: If each participant of a chess tournament plays exactly one
[#permalink]
Show Tags
11 Nov 2012, 05:07
derekgmat wrote: Thanks Karishma
I must admit I am struggling to work through your solution. I looked at the question again and answered it with the approach of working through each of the choices longhand.
If A, then 15 players means they each will play 14 games (so 14*15), divide the result by two for double counted games: 14*15 = 210 201/2 = 105, not 153, next
If B, 15*16 = 240 240/2 = 120, not 153, next
If C, 16*17 = 274 274/2 = 137, not 153, next
If D, 17*18 = 306 306/2 = 153 ANSWER D
Am I missing any important concepts by answering the question with this longhand method as opposed to your more structured approach?
Kind regards
Derek both are actually doing the same thing. You approach is a logical one while Karishma's is a structured one. But eventually both are doing same thing: n*(n1)/2 = 153 ( or 17*18/2 =153)
_________________



Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 9706
Location: Pune, India

Re: If each participant of a chess tournament plays exactly one
[#permalink]
Show Tags
11 Nov 2012, 19:40
derekgmat wrote: Thanks Karishma
I must admit I am struggling to work through your solution. I looked at the question again and answered it with the approach of working through each of the choices longhand.
If A, then 15 players means they each will play 14 games (so 14*15), divide the result by two for double counted games: 14*15 = 210 201/2 = 105, not 153, next
If B, 15*16 = 240 240/2 = 120, not 153, next
If C, 16*17 = 274 274/2 = 137, not 153, next
If D, 17*18 = 306 306/2 = 153 ANSWER D
Am I missing any important concepts by answering the question with this longhand method as opposed to your more structured approach?
Kind regards
Derek Your method is absolutely fine. Your logic of multiplying n players by (n1) games and dividing by 2 is great. The issue with your approach is that you need to calculate for every option. If the numbers were a little larger, you would end up doing calculations a number of times. For each option, you are doing this calculation (n1)*n/2 (A) 14*15/2 (B) 15*16/2 etc You are trying to find the option that will give you the result 153. I have done the same thing. I get (n1)*n/2 = 153. Instead of trying all options, you should multiply 153 by 2 to get 306 and then see which two numbers will end in 6. That way you will save a lot of calculations. 14*15  No 15*16  No 16*17  No 17*18  Yes 18*19  No As for the approach used by me to arrive at n*(n1)/2: Think of it this way  you make all participants stand in a straight line. The first one comes up and play a game with everyone else i.e. (n1) games and goes away. The next one comes up and plays a game with all remaining people i.e. (n2) people and goes away too. This goes on till last two people are left and one comes up, play a game against the other and they both go away. Hence, they end up playing (n1) + (n  2) + .... 3 + 2 + 1 games (total number of games) You must learn that sum of first m positive integers is given by m(m+1)/2 (very useful to know this) We need to sum first (n1) numbers so their sum will be (n1)n/2 Bunuel has used the combinatorics approach to arrive at n(n1)/2. There are n people and you want to select as many distinct two people teams as you can (since each person can play against the other only once) In how many ways can you do that? nC2 ways nC2 = n(n1)/2 So the calculation involved is the same in every case. The method of arriving at the equation is different.
_________________
Karishma Veritas Prep GMAT Instructor
Learn more about how Veritas Prep can help you achieve a great GMAT score by checking out their GMAT Prep Options >



Senior Manager
Joined: 13 Aug 2012
Posts: 401
Concentration: Marketing, Finance
GPA: 3.23

Re: If each participant of a chess tournament plays exactly one
[#permalink]
Show Tags
28 Dec 2012, 06:32
derekgmat wrote: If each participant of a chess tournament plays exactly one game with each of the remaining participants, then 153 games will be played during the tournament. Find the number of participants.
A. 15 B. 16 C. 17 D. 18 E. 19 \(\frac{P!}{2!(P2)!}=153\) \(\frac{P * (P1) * (P2)!}{(P2)!}=2*9*17\) \(P*(P1) = 18*17\) \(P = 18\) Answer: D
_________________
Impossible is nothing to God.



Intern
Joined: 05 Jun 2012
Posts: 23
WE: Marketing (Retail)

Re: If each participant of a chess tournament plays exactly one
[#permalink]
Show Tags
29 Dec 2012, 03:59
derekgmat wrote: If each participant of a chess tournament plays exactly one game with each of the remaining participants, then 153 games will be played during the tournament. Find the number of participants.
A. 15 B. 16 C. 17 D. 18 E. 19 Num of games  153, The total number of players that can be chosen as a pair would be nC2. Now look through the options & start form the middle number as the options are in ascending order  17C2 = 17*16/2 = 6 as last digit (17 *8) ignore Next number 18c2 = 18*17/2 = 3 as last digit this is the answer Choice D



Director
Joined: 23 Jan 2013
Posts: 525

Re: If each participant of a chess tournament plays exactly one
[#permalink]
Show Tags
01 Oct 2014, 04:32
x!/2!*(x2)!=153, so
Backsolving looks better:
get C=17 17!/2!*15!=17*16/2=136
go D=18 18!/2!*16!=18*17/2=153 (correct)
D



Manager
Joined: 05 Oct 2016
Posts: 90
Location: United States (OH)
GPA: 3.58

Re: If each participant of a chess tournament plays exactly one
[#permalink]
Show Tags
15 Sep 2017, 19:49
Bunuel wrote: derekgmat wrote: If each participant of a chess tournament plays exactly one game with each of the remaining participants, then 153 games will be played during the tournament. Find the number of participants.
A. 15 B. 16 C. 17 D. 18 E. 19 We are basically told that we can choose 153 groups of two players out of n players, thus \(C^2_n=153\) > \(\frac{n!}{(n2)!2!}=153\) > \(\frac{(n1)n}{2}=153\) > \((n1)n=306\) > \(n=18\). Answer: D. Similar questions to practice: http://gmatclub.com/forum/howmanydiag ... 01540.htmlhttp://gmatclub.com/forum/if10persons ... 10622.htmlhttp://gmatclub.com/forum/10businesse ... 26163.htmlhttp://gmatclub.com/forum/howmanydiff ... 29992.htmlhttp://gmatclub.com/forum/15chessplay ... 55939.htmlhttp://gmatclub.com/forum/thereare5c ... 27235.htmlHope it helps. bunuel how did you simply this equation?
_________________



Intern
Joined: 26 May 2017
Posts: 25

If each participant of a chess tournament plays exactly one
[#permalink]
Show Tags
15 Sep 2017, 21:36
derekgmat wrote: If each participant of a chess tournament plays exactly one game with each of the remaining participants, then 153 games will be played during the tournament. Find the number of participants.
A. 15 B. 16 C. 17 D. 18 E. 19 You can also approach the question with combinations. Let's say there are n people playing chess. You need to select 2 people from n people. Number of ways of doing that is nC2 = n!/(n2)!2! Reducing the above equation you get n(n1)/2 number of ways. Now, n(n1)/2 = 153 > n(n1) = 306 Try different options that fit the criteria. Quick method is to notice that the multiplication of n * n1 should give a units digit of 6. Answer D = 18, (18*17 = 306) Sent from my iPhone using GMAT Club Forum mobile app



Manager
Joined: 05 Oct 2016
Posts: 90
Location: United States (OH)
GPA: 3.58

Re: If each participant of a chess tournament plays exactly one
[#permalink]
Show Tags
16 Sep 2017, 15:34
craveyourave wrote: derekgmat wrote: If each participant of a chess tournament plays exactly one game with each of the remaining participants, then 153 games will be played during the tournament. Find the number of participants.
A. 15 B. 16 C. 17 D. 18 E. 19 You can also approach the question with combinations. Let's say there are n people playing chess. You need to select 2 people from n people. Number of ways of doing that is nC2 = n!/(n2)!2! Reducing the above equation you get n(n1)/2 number of ways. Now, n(n1)/2 = 153 > n(n1) = 306 Try different options that fit the criteria. Quick method is to notice that the multiplication of n * n1 should give a units digit of 6. Answer D = 18, (18*17 = 306) Sent from my iPhone using GMAT Club Forum mobile appplease simplify the above equation in more steps.
_________________



Senior SC Moderator
Joined: 22 May 2016
Posts: 3555

If each participant of a chess tournament plays exactly one
[#permalink]
Show Tags
16 Sep 2017, 16:45
SandhyAvinash wrote: craveyourave wrote: derekgmat wrote: If each participant of a chess tournament plays exactly one game with each of the remaining participants, then 153 games will be played during the tournament. Find the number of participants.
A. 15 B. 16 C. 17 D. 18 E. 19 You can also approach the question with combinations. Let's say there are n people playing chess. You need to select 2 people from n people. Number of ways of doing that is nC2 = n!/(n2)!2! Reducing the above equation you get n(n1)/2 number of ways. Now, n(n1)/2 = 153 > n(n1) = 306 Try different options that fit the criteria. Quick method is to notice that the multiplication of n * n1 should give a units digit of 6. Answer D = 18, (18*17 = 306) please simplify the above equation in more steps. SandhyAvinash , maybe this post will help. Here are a couple of sites I found that discuss the simplification of factorials with variables. . . This site is thorough and easy to understand: https://chilimath.com/lessons/intermediatealgebra/simplifyingfactorialswithvariables/This next site's treatment is shorter than that above, but I have seen others link to this site. (Other experts may have linked to chilimath as well. I ran the math as it was explained on chilimath. The math is fine.) http://www.purplemath.com/modules/factorial.htmHope they help! P.S. I think questions here work better if you type the symbol @ before the username in your post. The name will show up in blue, like this SandhyAvinash. If the person follows topics via email, s/he will get notified. S/he will not get notified with just SandhyAvinash, for example.
_________________
SC Butler has resumed! Get two SC questions to practice, whose links you can find by date, here.Instructions for living a life. Pay attention. Be astonished. Tell about it.  Mary Oliver



Manager
Joined: 05 Oct 2016
Posts: 90
Location: United States (OH)
GPA: 3.58

Re: If each participant of a chess tournament plays exactly one
[#permalink]
Show Tags
16 Sep 2017, 17:08
Thank you so much. I really appreciate your help sir. SandhyAvinash
_________________



Senior SC Moderator
Joined: 22 May 2016
Posts: 3555

If each participant of a chess tournament plays exactly one
[#permalink]
Show Tags
16 Sep 2017, 17:17
SandhyAvinash wrote: Thank you so much. I really appreciate your help sir. SandhyAvinashYou're very welcome. One last thing. When you "flag" someone by typing the @ symbol, use it before THEIR name (not yours). All these rules can get confusing. You are very polite, which I appreciate. Cheers!
_________________
SC Butler has resumed! Get two SC questions to practice, whose links you can find by date, here.Instructions for living a life. Pay attention. Be astonished. Tell about it.  Mary Oliver



Manager
Joined: 05 Oct 2016
Posts: 90
Location: United States (OH)
GPA: 3.58

Re: If each participant of a chess tournament plays exactly one
[#permalink]
Show Tags
17 Sep 2017, 05:45
genxer123 wrote: SandhyAvinash wrote: Thank you so much. I really appreciate your help sir. SandhyAvinashYou're very welcome. One last thing. When you "flag" someone by typing the @ symbol, use it before THEIR name (not yours). All these rules can get confusing. You are very polite, which I appreciate. Cheers! Oh thank you i got you genxer123
_________________



Current Student
Joined: 25 Jul 2011
Posts: 57
Location: India
Concentration: Strategy, Operations
GPA: 3.5
WE: Engineering (Energy and Utilities)

Re: If each participant of a chess tournament plays exactly one
[#permalink]
Show Tags
17 Sep 2017, 06:52
derekgmat wrote: If each participant of a chess tournament plays exactly one game with each of the remaining participants, then 153 games will be played during the tournament. Find the number of participants.
A. 15 B. 16 C. 17 D. 18 E. 19 It's a 10 second problem once you understand that the question is simply asking if nC2=153 then what is the value of n?consider only 2 players; A and B then no of matches "If each participant of a chess tournament plays exactly one game with each of the remaining participants"=1 (A vs B) as any particular match involves 2 players here..because each player plays against each of the remaining participants separately... so we only have to select 2 players...a simple combination.. to count a possible match from amongst the available players... If there are 2 players only...then selecting 2 out of 2=2C2=1 (AB) if there are 3 players only....then selecting 2 out of 3=3C2=3 (AB,AC,BC) if there are 4 players only....then selecting 2 out of 4=4C2=6 (AB,AC,AD,BC,BD,CD) Similarly... if there are n players only....then selecting 2 out of n=nC2=n!/2!(n2)! here..nC2=n!/2!(n2)!=153...put values back from options..we get n=18Further....had a match involved 3 players rather than 2....then the no. of matches would have been nC3... i.e., selecting 3 out of n players to play a match. Similarly....had a match involved 4 players rather than 2....then the no. of matches would have been nC4... i.e., selecting 4 out of n players to play a match. Thus, had a match involved r players rather than 2....then the no. of matches would have been nCr... i.e., selecting r out of n players to play a match....If each participant of a chess tournament plays exactly 2 game with each of the remaining participants=just multiply nC2 with 2..
_________________
Please hit kudos button below if you found my post helpful..TIA



Manager
Joined: 23 Apr 2018
Posts: 139

Re: If each participant of a chess tournament plays exactly one
[#permalink]
Show Tags
21 Aug 2019, 09:54
VeritasKarishma wrote: derekgmat wrote: Thanks Karishma
I must admit I am struggling to work through your solution. I looked at the question again and answered it with the approach of working through each of the choices longhand.
If A, then 15 players means they each will play 14 games (so 14*15), divide the result by two for double counted games: 14*15 = 210 201/2 = 105, not 153, next
If B, 15*16 = 240 240/2 = 120, not 153, next
If C, 16*17 = 274 274/2 = 137, not 153, next
If D, 17*18 = 306 306/2 = 153 ANSWER D
Am I missing any important concepts by answering the question with this longhand method as opposed to your more structured approach?
Kind regards
Derek Your method is absolutely fine. Your logic of multiplying n players by (n1) games and dividing by 2 is great. The issue with your approach is that you need to calculate for every option. If the numbers were a little larger, you would end up doing calculations a number of times. For each option, you are doing this calculation (n1)*n/2 (A) 14*15/2 (B) 15*16/2 etc You are trying to find the option that will give you the result 153. I have done the same thing. I get (n1)*n/2 = 153. Instead of trying all options, you should multiply 153 by 2 to get 306 and then see which two numbers will end in 6. That way you will save a lot of calculations. 14*15  No 15*16  No 16*17  No 17*18  Yes 18*19  No As for the approach used by me to arrive at n*(n1)/2: Think of it this way  you make all participants stand in a straight line. The first one comes up and play a game with everyone else i.e. (n1) games and goes away. The next one comes up and plays a game with all remaining people i.e. (n2) people and goes away too. This goes on till last two people are left and one comes up, play a game against the other and they both go away. Hence, they end up playing (n1) + (n  2) + .... 3 + 2 + 1 games (total number of games) You must learn that sum of first m positive integers is given by m(m+1)/2 (very useful to know this) We need to sum first (n1) numbers so their sum will be (n1)n/2 Bunuel has used the combinatorics approach to arrive at n(n1)/2. There are n people and you want to select as many distinct two people teams as you can (since each person can play against the other only once) In how many ways can you do that? nC2 ways nC2 = n(n1)/2 So the calculation involved is the same in every case. The method of arriving at the equation is different. using your approach, we can instead save time and see which units digit gives 3 in the end.. 17*9 gives a 3 (dividing by 2 before multiplying the bigger terms) therefore 18 (D) is the answer




Re: If each participant of a chess tournament plays exactly one
[#permalink]
21 Aug 2019, 09:54






