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The limit really means 0+ in this case so the LHL will not come into play as someone pointed out. Therefore the answer will be 0.
Venkat, I'm confused as to why you say the function is not defined after 2? Maybe I've read things wrongly
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I'm not sure what the original problem was, but your solution seems correct, assuming the first step is correct.
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ppl.
please do not waste time trying to do stuff with the goldbach conjecture. once you get into iit you will have enough time :)
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All these questions are similar in nature. Just use the fact that (a+ib)(a-ib)=a^2+b^2
in the denominator
so (2+i)^2/(3+i) becomes (2+i)^2*(3-i)/10 and then you can simplify this further.
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catch_arrnie is on the right track, so too nano
the expansion idea is great but I do think it needs to be explained a bit
7^256 ... we have to consider the last 3 digits of this number and not just the last digit.
Now 7^4 = 2401, so 7^256 will have the same last 3 digits as 401^256. Expanding 401^256 as (400+1)^256, we can see that the last 3 digits will
be 601
(7^255)*256*10 .. we need to find the last 3 digits of this number.
we know the last digit will be 0. For 7^255 = 343*(7^252), we use the same idea for 7^252 and show that the last 3 digits will be 201. multiplying by 343 we get 943 as the last 3 digits. multiply this by 256 and we get ...408. Finally multiply this by 10 and we get the last 3 digits of this term to be 080
and as nano points out, the third term is not significant here because 255*256 will result in an extra 0 and hence a term with 3 zeroes.
final answer therefore is 080+601 = 681
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This seems like a very interesting problem
Let me make sure I understand the question correctly
let's say the Rs 50 people are named A B C D E
and the Rs 100 ppl are G H I J K
then will we consider A B C D E G H I J K (the order of arrival) as one solution and
B C A D E G H I J K as another one? or will they be counted the same because we are treating the Rs 50 people as indistinguishable?
thanks for the clarification
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Ok, I'll be honest and admit I don't see any answer to this question right away. It seems to me like some kind of manipulation will be needed after repeated substitution ...
but this looks very much like the equation for fibonacci numbers and they are not periodic, so I'm having some trouble thinking of this function as being periodic.
Do you know the answer by any chance? In that case I might get a clue and be able to help you out.
Sorry about my inability to solve it
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you can also do this by converting it into sin/sin form.
so cos 3x = -sin 3(pi/2-x)
cos 7x = -sin 7(pi/2-x)
and it is easy to see that the limit as y goes to 0 of sin Ny/ sin My will be N/M
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there has to be some trick involved in these questions, otherwise i'd just write computer programs for them :).
Anyway, here's a solution for the division by 1 to 20
Let's first see why 2520 worked out for the 1 to 10 example
Begin by multiplying out all the primes from 1 to 10, so that is 2*3*5*7=210
So we've taken care of 2,3,5,6,7,10 by doing this.
To take care of 4 ... we need to multiply by 2 ..so now our number is 420
To take care of 8.. multiply by 2 again ... 840
Finally to take care of 9 .. multiply by 3 and end up with 2520
For 1 to 20 the same logic can be applied
first 2*3*5*7*11*13*17*19 = 9699690
numbers taken care of ... 1,2,3,5,6,7,10,11,13,14,15,17,19
for 4 .. multiply by 2 = 2*3*5*7*11*13*17*19*2
numbers taken care of ... 1,2,3,4,5,6,7,10,11,12,13,14,15,17,19,20
for 8 ..multiply by 2 = 2*3*5*7*11*13*17*19*2*2
numbers taken care of ... 1,2,3,4,5,6,7,8,10,11,12,13,14,15,17,19,20
for 9 .. multiply by 3 = 2*3*5*7*11*13*17*19*2*2*3
numbers taken care of ... 1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,17,18,19,20
only 16 is left, take care of that by multiplying by 2
2^4*3^2*5*7*11*13*17*19
232792560
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The correct responses have been posted by vishak and nick. good job guys.
Basically (n-2) triangles for an n sided polygon.
Also for the hexagon, the interior angles are all 120 deg, this should help you come up with the solution.
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great job Nick!
I don't think there are any shortcuts here apart from observing that for a lot of letters the number of words that you can form starting with that letter will be the same.
There might be some nice result somewhere, but I don't think it is worth your time to a) find it and b) memorize it.
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LH works but here is another method
(1 + cos \pi x) / tan^2 \pi x
(1 + cos \pi x) * cos^2 \pi x / sin^2 \pi x
multiply and divide by ( 1 - cos \pi x) and simplify to get the result of 1/2
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It is mainly related to the analysis and design of things that you would like a computer to do. So yes, it is mostly related to software, but it is more to do with the design than with the implementation.
To give you a real quick and dirty example, if you join Comp Sc you will learn more about how to do sorting than about how to write a sorting program in Java.
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the problem with this solution seems to be situations like
when a=2 b=2 c=3 is also a valid isoceles triangle. this is not accounted for anywhere. i tried doing it this way initially and things got messy which is why the other approach was taken.
also equilateral triangles are isoceles. this is a minor fix however.
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well consider it this way
let the equal sides be of length 1 .. then we have only one choice for the non-equal side, that would be 1. Since if we take 2 then triangle inequality will mess things up
if equal sides are of length 2 .. third side can either be 1 or 2 or 3... basically until we hit 2*2.
if equal sides are of length 3 .. third side can be 1 or 2 or 3 or 4 or 5.
now this logic continues until we hit 998 ..since 2*998=1996 we can basically pick anything for the third side from 1 to 1994
So it looks like the sum should be 1+3+5+...1993+ 997*1994
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Any pair in which the order is important. It is typically used for sets and relations.
For example if you take the cartesian product of two sets X and Y. Then a member of this set is specified by (x,y) where the order becomes important because x comes from X and y comes from Y.
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There are lots of different ways of thinking of 'e'. From a practical point of view, it makes it so much easier to talk about certain infinite series. Historically, it seems to have come about due to attempts at computing the limit of (1+1/n)^n as n grew larger and larger.
While it would have been nice if 'e' had a cool physical interpretation like  does (circumference/dia etc etc), I cannot recall such a thing.
e^(i \theta) is not just notation, it means a lot in the theory of complex numbers. With this definition you can do things like take log of complex numbers and stuff like that. So for example, the annoying question of what is log(-1) can now be answered :).
I hope this clears some of your doubts. In truth 'e' might've historically come about to solve some specific problem, but it surprisingly showed up in other contexts too and now mathematicians have embraced it as one of the fundamental constants.
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This can also be done by taking log of the (n!)^(1/n)
in which case we get
(1/n)log(n)+(1/n)log(n-1) + ...
and we can compute the limit of each term.
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I tried this with IE and it worked. You can attach files which are jpegs etc. So can take your pdf and make it into a jpg/gif and then attach it.
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The attachment did not work in firefox but I'm trying it now with IE.
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