Simple math problem for the late night crew
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Date: December 15th, 2017 4:54 AM Author: Fiercely-loyal Exhilarant Background Story Public Bath
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 5:04 AM Author: Emerald jap state
in my head:
edit:
n=35
(m+k)/2 = 18
(http://www.autoadmit.com/thread.php?thread_id=3829848&forum_id=2#34925148) |
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Date: December 15th, 2017 5:24 AM Author: free-loading splenetic whorehouse gaming laptop
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.
(http://www.autoadmit.com/thread.php?thread_id=3829848&forum_id=2#34925210) |
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Date: December 15th, 2017 5:32 AM Author: free-loading splenetic whorehouse gaming laptop
???
economics/finance expert witness for damage$$$?
friend of mine from ug works as a consultantttt at a shop that does that shit. seems chill.
(http://www.autoadmit.com/thread.php?thread_id=3829848&forum_id=2#34925227) |
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Date: December 15th, 2017 5:47 AM Author: Emerald jap state
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
(http://www.autoadmit.com/thread.php?thread_id=3829848&forum_id=2#34925253) |
Date: December 15th, 2017 5:21 AM Author: free-loading splenetic whorehouse gaming laptop
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.
(http://www.autoadmit.com/thread.php?thread_id=3829848&forum_id=2#34925199) |
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Date: December 15th, 2017 6:09 AM Author: Purple incel
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
(http://www.autoadmit.com/thread.php?thread_id=3829848&forum_id=2#34925314)
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Date: December 15th, 2017 5:38 AM Author: Purple incel
2plus 2 is 4
Minus 1 that’s 3
Quick mafs
(http://www.autoadmit.com/thread.php?thread_id=3829848&forum_id=2#34925239) |
Date: December 15th, 2017 8:48 AM Author: contagious henna bawdyhouse famous landscape painting
#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
(http://www.autoadmit.com/thread.php?thread_id=3829848&forum_id=2#34925665) |
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Date: December 15th, 2017 10:23 AM Author: contagious henna bawdyhouse famous landscape painting
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.
(http://www.autoadmit.com/thread.php?thread_id=3829848&forum_id=2#34926287) |
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Date: December 15th, 2017 1:31 PM Author: Fiercely-loyal Exhilarant Background Story Public Bath
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)
(http://www.autoadmit.com/thread.php?thread_id=3829848&forum_id=2#34927693) |
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Date: December 15th, 2017 10:57 AM Author: Charcoal vengeful locale
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
(http://www.autoadmit.com/thread.php?thread_id=3829848&forum_id=2#34926521) |
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Date: December 15th, 2017 6:08 PM Author: Charcoal vengeful locale
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
(http://www.autoadmit.com/thread.php?thread_id=3829848&forum_id=2#34930017) |
Date: December 15th, 2017 10:19 AM Author: Histrionic Travel Guidebook Volcanic Crater
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
(http://www.autoadmit.com/thread.php?thread_id=3829848&forum_id=2#34926251) |
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