mathmandan
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Task 1, ><>, 10 + 3 = 13 bytes ::*-:*4,n; Run this using the official Python interpreter using the -v flag (at a cost of 3 bytes). This squares the quantity (n - n*n) and divides by 4, which of ...

Python 3, 40 38 bytes Thanks to squid for saving 2 bytes! lambda n:n in(i*i for i in range(n+1)) Try it online! Too slow to return an answer for 2147483647 in a reasonable amount of time. (But ...

Python 2, 104 bytes N=input();i=s=0;l=() while N!=sum(l):exec'while sum(l)'+['<N:i+=1;l+=i,','>N:l=l[1:]'][s%2];s+=1 print s Try it online! Alternates between adding terms to the end of the ...

Python 2, 77 79 bytes a=input();u=[];j=1 for x in a: u+=[[x,j][x in u]] while j in u+a:j+=1 print u Takes keyboard input, like [3, 3, 3, 3, 3, 3]. Just keep track of the smallest positive integer ...

Python 2, 99 86 bytes Thanks to @Dennis for several improvements totaling 13 bytes! s=0,1,2,3 exec't=1\nwhile t==s[-1]or s.count(t)/s[t]:t+=1\ns+=t,;'*input() print s[-4] The program proceeds pretty ...

Python 2, 26 t=lambda n:len(bin(n+n))-4 This is similar to the Python 3 answer by Tim S. However, doubling n and then subtracting 4 from the length has the advantage of working whether n is ...

Python 2, 101 bytes N=input();r=range for i in r(N,0,-1):print''.join('# '[i>sum(x%-~p<1for p in r(x))]for x in r(1,1+N)) Try it online! This produces (a lot of) vertically leading whitespace....

Python 2, 99 bytes a,b=input();o=0;p=1-2*(a*b<0);a,b=abs(a),abs(b) while a:o+=a%10*(b%10)*p;p*=100;a/=10;b/=10 print o A lot of the bytes are there to account for the sign in case of negative ...

Python, 50 bytes def f(a): for l in a[1:-1]:l[1:-1]=[0]*(len(l)-2) Accepts a list of lists, and modifies it in place. Python's slice syntax is not inconvenient for this task. I learned that ...

Python 2, 264 bytes There's already a remarkable Python 3 answer by mbomb007. This answer steals flagrantly from that one (in particular the "Alex is wrriognhgt" trick). And this answer is ...

Python 2, 67 This uses the same summation as Anthony Roitman, and the same recursion as Morgan Thrapp. f=lambda l,n:f([sum(l[:i+1])for i in range(len(l))],n-1)if n else l I developed this solution ...

Python2 - 37 g=lambda x:(x%4or x%400and x%100<1)<1 Note that if a is a nonnegative integer, then a<1 is a short way of writing not bool(a). The last <1 thus effectively converts the ...

Python 2, 113 bytes r=range lambda N:[n for n in r(1,N)if 1-any((bin(k).count('1')<3)*all((n^k)%q for q in r(2,n^k))for k in r(n+1))] (The second line is an unnamed function which returns a list ...

Python 2, 111 bytes def p(n,b=0,o=''): if n<1:return n for c in str(n):o+=c+'**'[b:];b=~b j=eval(o+'1');return p(j)if j-n else j The idea is to make a string where the digits of n are ...

><>, 21 + 3 = 24 bytes, 22 points :0)$:0($:1\$2%2*-*+-n; Use the official Python interpreter, and use the -v command line option to enter input, at a cost of 3 bytes. I have a feeling that ...

Python 2, 59 bytes def p(l): print l;o=ord(l[0])-97 if o:p(l[o::-1]+l[o+1:]) I guess this is a rather straightforward answer. Uses recursion and Python's slice syntax. Call as: p('eabdc').

Python 2, 96 bytes def g(s):u=7<len(s)<10and sum(x*int(('0'+s)[-x])for x in range(2,10))-int(s[-1]);print(u%11<1)*u Takes a string as input. The function adds a '0' to the front of the ...

Python 2, 123 121 bytes from random import* def f(a,b,c): while c*b: r=randint(1,20);print r;c-=r<a;b-=r>=a if r in[1,20]:return r>9 return c (This answer mixes spaces and tabs, so ...

Python 2, 127 bytes N=input();a,b=0,1 while N:a,b=b,a+b;t=int(''.join(c for c in hex(b)if ord(c)<65));N-=(t>1)*all(t%x for x in range(2,t)) print b The algorithm could be a lot more efficient. ...

Python 2, 175 - 5 = 170 bytes n=input();s='';u=j=-1 while n>len(s): while u>2:u=int(s[0]);s=s[1:] u+=1;t=u+s;m=c=0 while t[c:]*0**m:c+=1;i=t[c:].find(t[:c]);m=j<i<=c if c>=len(t):...

Python 2, 168 bytes n=input();e=[-1,21,38,62,80,108,140,207];m=1 while n>e[m]:m+=1 print[n-e[m-1]+' '+'Qupu Blinkorp Paas Karpasus Dumaflop Lindilo Fwup'.split()[m-1],'Floopdoor'][n==80] This ...

Python 2.x, 139 bytes by Puzzled H(2) and H(128) both return 16645614427504350476847004633262883518.

Python 2, 44 43 bytes (Crossed out 44 is still 44.) Thanks to Fireflame241 for saving a byte! P=input();i=P/3 while i*10%P-1:i-=1 print i Try it online! There is exactly one number between 0 and P-1 ...

Python 2, 87 bytes a=input() t=int(len(a)/2.71828) m=max(a[:t]+[-1]) for x in a[t:]: if x>m:break print x The user enters the array as a list, with square brackets and commas. Python 2's input() ...

Python 2, 71 bytes def a(n,x=0,y=1,z=2,w=1,p=0): if~n:print[x,z][p];a(n-1,y,x+y,w,z+w,~p) This definitely seems too long. However, I was pleased that I got to use the bitwise not operator...twice. ...

Python 2, 67 62 bytes m=lambda x,y:y and x+m(x,y-1);p=lambda x,y:y<1or m(x,p(x,y-1)) This is an obvious solution: define multiplication in terms of addition, then exponentiation in terms of ...

Python 3, 69 bytes No one has posted a Python solution yet, so here's a fairly straightforward implementation of a "chunkiness" function. It short-circuits on a string of length 1, and ...

Python2 Update: New version doesn't use any mod or division operations. word_dict = {3: 'Fizz', 4: 'Jazz', 5: 'Buzz'} def fizz_jazz_buzz(n, d): counters = {k: k for k in d} for i in xrange(...