definitely, haha !
Prime factor of 105 -> 3*5*7
The number has 3, 5, 7, and 1
These are 4 distinct, so there are 4! ways
this is the only case b/c there is no multiplication of 3,5,7 that is 1 digit
The answer is 4!
note 1 = 3/3
2/3, 3/3, 4/3 etc.
(2 + 7*1)/3 = 9/3 = 3
This problem is controversial. The shape isn't even a parallelogram, so none.
Thank you, catmg! :)
301 + 303 + 305 + 307 + … + 497 + 499
301 * 100 + 2(1+2+3+4+…+98+99)
30100 + 2(99)(100)/2
30100 + 9900
40000
6! / (1! 2! 3!) = (6*5*4)/(2)=6*5*2=60
Looking at the AoPS Honor Code here: https://artofproblemsolving.com/school/handbook/current/honorcode which seems to pop up every class that starts for you, it clearly says that "You should not search for potential solutions to problems online or elsewhere. (Especially searching for the problem statement!) Using the internet in this way to search for answers is against our Course Policies! Instead, if you need some assistance with a particular problem, we welcome you to ask questions on your course message board. Your instructor and other AoPS staff members will be happy to help. Again, the solution that you submit should be your own solution in your own words. If you copy a solution from the internet into your submission, this is misrepresenting another person's work as your own. If you are reading a solution from the internet as you create your submission, this is misrepresenting another person's work as your own." So no.
Troll solution: https://www.desmos.com/calculator/fyaevo8phf shown on the graph, asymptote is x=2 . Y infinitely goes up and when y is infinitely up x will be undefined
Thus when y = busy beaver ^ busy beaver … (busy beaver amount of busy beavers in the exponent which is exponented by) x will be undefined.
Dunno about vectors, but in trig we have that sin^2(theta) + cos^2(theta)=1 which is like the pythog theorem
