The most prestigious law school admissions discussion board in the world.
|
 |
 
|
Simple math problem for the late night crew
Poast new message in this thread
Date: December 15th, 2017 4:54 AM Author: slap-happy dopamine foreskin Integers 1, 2, 3, 4, 5,...,n, where n > 2, are written on a board. Two numbers m, k such that 1 < m < n, 1 < k < n are removed and the average of the remaining numbers is found to be 17. What is the maximum sum of the two removed numbers? (http://www.autoadmit.com/thread.php?thread_id=3829848&forum_id=2#34925130) |
Date: December 15th, 2017 5:00 AM Author: Wonderful piazza son of senegal I don't know how to do this (http://www.autoadmit.com/thread.php?thread_id=3829848&forum_id=2#34925140) |
|
Date: December 15th, 2017 5:04 AM Author: Dull library the fuck (http://www.autoadmit.com/thread.php?thread_id=3829848&forum_id=2#34925147) |
|
Date: December 15th, 2017 5:04 AM Author: Indigo gaming laptop in my head:
edit: n=35 (m+k)/2 = 18 |
|
Date: December 15th, 2017 5:08 AM Author: Indigo gaming laptop so what’s the answer? (http://www.autoadmit.com/thread.php?thread_id=3829848&forum_id=2#34925151) |
 |
Date: December 15th, 2017 5:21 AM Author: slap-happy dopamine foreskin see below so i can edit (http://www.autoadmit.com/thread.php?thread_id=3829848&forum_id=2#34925202) |
|
Date: December 15th, 2017 5:23 AM Author: Indigo gaming laptop fuck I used to be able to do shit like this in my head, arteries must be hardening up *cue wlmas post about me being a, what was it, “pants shitting boomer”* |
|
Date: December 15th, 2017 5:24 AM Author: charcoal autistic indian lodge jap did u cop 800 SAT math?
i've never been able to do problems like this "in my head" but have always enjoyed set theory and number theory problems.
when i used to tutor kids on the SAT i'd tell them the SAT math section was a logic test, not a math test. |
|
Date: December 15th, 2017 5:27 AM Author: Indigo gaming laptop yeah and for GRE too agreed about the logic test, that’s the “doing it in the head” part, can’t do long divison in my head like true high functioning autists can |
|
Date: December 15th, 2017 5:31 AM Author: charcoal autistic indian lodge jap often times i'd end up doing a lot of the calculations in my head because it was just faster, but i always wanted to sketch out the 'parameters' of the problem on note paper to help me work through the thought process. this was for 'hard' SAT problems and for math olympiad and similar contests.
i copped 800 SAT math, but never took the GRE. 800 GRE math seems preftigif though. i took the GMAT senior year and barely copped a 49Q (88th percentile) math score, because actual indian turds from turdistan ruin the quant section for everyone. overall score was good though. |
|
Date: December 15th, 2017 5:34 AM Author: Indigo gaming laptop the studybots from India and China who prepare for these tests for years in advance skew the whole thing a LOT (http://www.autoadmit.com/thread.php?thread_id=3829848&forum_id=2#34925231) |
|
Date: December 15th, 2017 5:38 AM Author: charcoal autistic indian lodge jap when i took it getting two subscore points off a perfect quant score (49/51) equated to 85th - 90th percentile depending on the test.
now i've heard a perfect Q score (51/51) is at the 91st percentile due to mainland NOWIGs buttfucking the GMAT curve. (yet they continually score 45th percentile on verbal LJL) |
|
Date: December 15th, 2017 5:26 AM Author: slap-happy dopamine foreskin if you could do shit like this in your head then you wouldnt be a lolyer. i call flame (http://www.autoadmit.com/thread.php?thread_id=3829848&forum_id=2#34925214) |
|
Date: December 15th, 2017 5:28 AM Author: Indigo gaming laptop I don’t work as a lolyer (although I have a license ljl), I whore out my PhD for money to lolyers (http://www.autoadmit.com/thread.php?thread_id=3829848&forum_id=2#34925221) |
|
Date: December 15th, 2017 5:30 AM Author: slap-happy dopamine foreskin ok then its possible. because people who can solve this even with pencil and paper (forget in the head) at HS level are going to be doing regional math olympiads and pursuing STEM degrees/careers (http://www.autoadmit.com/thread.php?thread_id=3829848&forum_id=2#34925223) |
|
Date: December 15th, 2017 5:33 AM Author: Indigo gaming laptop not claiming I could have done this one, but this kind of problem — once upon a time fuck those hardening arteries! |
|
Date: December 15th, 2017 5:32 AM Author: charcoal autistic indian lodge jap ???
economics/finance expert witness for damage$$$?
friend of mine from ug works as a consultantttt at a shop that does that shit. seems chill. |
|
Date: December 15th, 2017 5:35 AM Author: Indigo gaming laptop that general kind of thing (http://www.autoadmit.com/thread.php?thread_id=3829848&forum_id=2#34925233) |
|
Date: December 15th, 2017 5:38 AM Author: charcoal autistic indian lodge jap do u like it? good $$? are you able to mostly set your own hours or do you feel beholden to clients most days? (http://www.autoadmit.com/thread.php?thread_id=3829848&forum_id=2#34925238) |
|
Date: December 15th, 2017 5:47 AM Author: Indigo gaming laptop I’m sincerely thinking about what the real answer is
yeah it’s good to get paid for being “smart” and chargeout rates are higher than lolyer partners get, which pisses them off sometimes ljl
I’m still beholden to clients though, and because lolyers are always in a constant state of panic about what their clients think, that influences me too and sometimes makes me the lawyer’s bitch
but I have an inbuilt defense mechanism in that I can only say what is defensible and can’t make shit up otherwise I’ll get torn apart, which gives me a nice out if a lawyer wants me to do shit I can’t defend
but in summary everybody is somebody else’s bitch and I won’t weep when I decide I’ve had enough |
|
Date: December 15th, 2017 5:50 AM Author: charcoal autistic indian lodge jap how are your hours?
honestly job sounds pretty "chill" compared to a (((tech))) firm that's trying to 'change the world' |
|
Date: December 15th, 2017 5:55 AM Author: Indigo gaming laptop hours depend completely on what I have to do I have slept only about 8 hours in the last 3 nights: part of that is because of a shitty redeye flight, but most of that is because of sitting in a hotel room working through a couple of nights finishing a report for submission to a decision maker, and I will be working all weekend but then I probably won’t go near an office for the next 6 weeks after this is done and I send them a fat bill |
|
Date: December 15th, 2017 6:08 AM Author: slap-happy dopamine foreskin how are the travel allowances? can you fly business class, stay at $500 dollar a night hotels and expense it with no questions asked? if so then its 180 even if you have to travel a lot (http://www.autoadmit.com/thread.php?thread_id=3829848&forum_id=2#34925310) |
|
Date: December 15th, 2017 6:16 AM Author: Indigo gaming laptop mostly, but I try not to be unreasonable about it honestly though, business travel is work not a vacation, and a cheese sandwich at home is almost always a better meal than anything while traveling but nice hotels etc makes it a lot easier |
|
Date: December 15th, 2017 5:14 AM Author: slap-happy dopamine foreskin EDITED (http://www.autoadmit.com/thread.php?thread_id=3829848&forum_id=2#34925177) |
|
Date: December 15th, 2017 5:21 AM Author: charcoal autistic indian lodge jap seems like a fun set problem.
i've had too much (((poison))) tonight but i will revisit tomorrow and scratch out an answer. try to avoid spoilers/solutions til then. |
|
Date: December 15th, 2017 6:09 AM Author: Talented fiercely-loyal jew It’s not a set problem. It’s arithmetics.
Obviously m=n-1 and k=n-2
Average of 1...n is (1+n)/2
Total without m,k is (1+n)*n/2-3 or 17*(n-2) So
(n^2+n)/2=17n-34+3=17n-31
N^2+n = 34n-62
N^2-33n+62=0
Where did I fuck up
|
|
Date: December 15th, 2017 9:52 AM Author: Deranged pervert locale "Obviously m=n-1 and k=n-2"
I made the same mistake below. In fact we don't know if m _can_ be n-1 or n-2, because there's the constraint that the average of the sum 1...n -(m+k) must be 17. |
|
Date: December 15th, 2017 5:38 AM Author: Talented fiercely-loyal jew 2plus 2 is 4 Minus 1 that’s 3 Quick mafs |
|
Date: December 15th, 2017 5:56 AM Author: floppy bawdyhouse mental disorder Fuck I messed up one of the equations (http://www.autoadmit.com/thread.php?thread_id=3829848&forum_id=2#34925272) |
|
Date: December 15th, 2017 6:48 AM Author: Talented fiercely-loyal jew The answer is 15.5 (http://www.autoadmit.com/thread.php?thread_id=3829848&forum_id=2#34925403) |
|
Date: December 15th, 2017 8:48 AM Author: Deranged pervert locale #QUESTION: Integers 1, 2, 3, 4, 5,...,n, where n > 2, are #written on a board. Two numbers m, k such that 1 < m < n, 1 #< k < n are removed and the average of the remaining #numbers is found to be 17. What is the maximum sum of the #two removed numbers?
Assume k>m>n (order of k, m doesn't matter)
Two removed numbers max sum, so removed numbers are n-1, n-2 ("they've got to be in the last slot") hence k+m = n + (n-3) = 2n-3 Sum of #s from 1 to n is n(n+1)/2 n^2+n/2 - (2n-3) is the numerator for the mean = to 17 n-2 is the denominator then i remembered idk algebra working by inspection, n has got to be 35 (SAT math style just double 17 then inspect nearby averages with penultimate and antepenultimate #s removed) 35*2=70-3=67 ans=67 |
|
Date: December 15th, 2017 8:50 AM Author: crystalline office "Assume k>m>n (order of k, m doesn't matter)"
dingfag |
|
Date: December 15th, 2017 8:53 AM Author: Deranged pervert locale thks. I guess it says max sum but it is wrong to suppose therefore they've got to be the last two numbers before n, the reason being the scale to 17 might vary. (http://www.autoadmit.com/thread.php?thread_id=3829848&forum_id=2#34925683) |
|
Date: December 15th, 2017 9:35 AM Author: Deranged pervert locale Okay, m = 2, k=n-1.
n is 33. hence the answer is 32+2=34. |
|
Date: December 15th, 2017 9:42 AM Author: Deranged pervert locale [reasoning - yields 17 avg and you can see geometrically that adjusting by altering the "2" wouldn't yield a diff number b/c you still need to avg to 17 so you'd need to adjust the other side]
[to elaborate, the pertinent multiple is 34, but you need to add 1, so start w/ 33 for n. the only way to preserve the property of averaging to 17 is to take from both sides of {2...32} equally. hence m+k=34.] |
|
Date: December 15th, 2017 10:21 AM Author: Slate underhanded site personal credit line To max m + K you can apply same reasoning for n=34 and get 51 (http://www.autoadmit.com/thread.php?thread_id=3829848&forum_id=2#34926259) |
|
Date: December 15th, 2017 10:23 AM Author: Deranged pervert locale Doesn't work out to avg 17, no?
(1+3+4+5+6+7+8+9+10+11+12+13+14+15+16+17+18+19+20+21+22+23+24+25+26+27+28+29+30+31+32+34)/32 = 17.5.
Reason is you're adding in 1/2 due to the extra 1 at the beginning of the sequence. |
|
Date: December 15th, 2017 10:54 AM Author: Deranged pervert locale n/m u r right, thank you! i was on the right track but did not close the pincer on mid #'s. ;_; I suck at maths.
|
|
Date: December 15th, 2017 10:58 AM Author: Slate underhanded site personal credit line See my reasoning below. If you do it your way, you just had to take out larger numbers to get the average down to 17 (http://www.autoadmit.com/thread.php?thread_id=3829848&forum_id=2#34926524) |
|
Date: December 15th, 2017 1:31 PM Author: slap-happy dopamine foreskin you got very close but answer is wrong.
you need to also consider the other side k=2,m=3. this is when the average of remaining numbers will be maximum
with the two equations (k=n-1,m=n-2 && k=2,m=3) you can solve for n and get a range
after that it is trivial
correct answer is 51 (n=34) |
|
Date: December 15th, 2017 1:32 PM Author: Deranged pervert locale yea figured it out in interim.
##EDIT oic ty for your cmt. I'm really surprised I got as far as I did. |
|
Date: December 15th, 2017 9:04 AM Author: Slate underhanded site personal credit line I got m+k = 51 where n is 34 (http://www.autoadmit.com/thread.php?thread_id=3829848&forum_id=2#34925717) |
|
Date: December 15th, 2017 10:57 AM Author: Slate underhanded site personal credit line Showing my work: (sum of (1 to n) - (m+k) ) divided by remaining numbers = 17
So basically, insert Gauss sum: (n*(n+1))/2 - (m+k) = (n-2)*17
m+k = (n*(n+1))/2 - (n-2)*17
Solve for m+k using various n's, and then make sure that m+k make sense: For n = 33: m+k = 33*34/2 - 31*17 = 34 For n = 34: m+k = 34*35/2 - 32*17 = 51 For n = 35: m+k = 35*36/2 - 33*17 = 69
However, there is no such combo of m < n and k < n where n = 35 such that m+k = 69.
So max m+k = 51, for n = 34 |
|
Date: December 15th, 2017 10:59 AM Author: Deranged pervert locale Thanks for poasting, helpful to see your setup. (http://www.autoadmit.com/thread.php?thread_id=3829848&forum_id=2#34926534) |
|
Date: December 15th, 2017 1:25 PM Author: Deranged pervert locale I noticed the valid sums (34, 51) are integer multiples of 17, whereas the invalid sum (69) is off by 1.
This suggests to me there's some sorta geometric intuitive way to do this problem, which obviates the need for any bruteforcing - wish I could spend the day looking @ this problem & similar ;_;. |
|
Date: December 15th, 2017 6:08 PM Author: Slate underhanded site personal credit line Yeah for any stated avg a (in this case, a=17), the answer for max m+k will be 3a
And to clarify, 34 and 51 aren't the only valid sums. Using same formula above (which actually comes out to 0.5n^2-16.5n+34): Less than 31, m+k is negative For n = 31, m+k=3 For n = 32, m+k=18
https://www.wolframalpha.com/input/?i=0.5n%5E2-16.5n%2B34 |
|
Date: December 15th, 2017 6:17 PM Author: Deranged pervert locale Many thanks! (http://www.autoadmit.com/thread.php?thread_id=3829848&forum_id=2#34930076) |
|
Date: December 15th, 2017 1:32 PM Author: slap-happy dopamine foreskin WINNER. congrats (http://www.autoadmit.com/thread.php?thread_id=3829848&forum_id=2#34927700) |
|
Date: December 15th, 2017 1:32 PM Author: Deranged pervert locale (http://www.autoadmit.com/thread.php?thread_id=3829848&forum_id=2#34927704) |
|
Date: December 15th, 2017 1:34 PM Author: Deranged pervert locale Thanks for poasting, german idealism, fun problem. (http://www.autoadmit.com/thread.php?thread_id=3829848&forum_id=2#34927717) |
|
Date: December 15th, 2017 10:19 AM Author: fragrant canary reading party Didn't solve it yet but here are my thoughts (yes I'm a dumb):
"Integers 1, 2, 3, 4, 5,...,n, where n > 2, are written on a board. Two numbers m, k such that 1 < m < n, 1 < k < n are removed and the average of the remaining numbers is found to be 17. What is the maximum sum of the two removed numbers?"
Assume "p" is the sum off all the integers as described. Therefore: (p-m-k) / (n-2) = 17. This is given in the problem. n cannot be 3 because n > m and k >1. and n>2. n can be 4 at the smallest. this would make m and k 3 and 2 at the minimum.
What fractions give us 17? 17/1, 34/2, 51/3, 68/4, 85/5, 102/6, etc. 17/1 is not possible because n is at least 4. n must also be even because n-2 with n at least 4 will always be even. so 51/3, 85/5, etc is out.
therefore p has to be at least 40: (40-3-2) / (4-2) = 17.
p the sum of all integers is at least 40. n is at least 4. n - 2 = some even number
im stuck?
EDIT: how do we get to 40 adding consecutive integers and keeping n = 4? sum 1 - 4 |