Okay um...heres the solution to my problem
[t] [e] [n] [h] [o] [r] [s] [e] [s]
More of a riddle actually 
This is just an excuse for people to post their homework and have people do it for them... wish I had thought of this when I still did maths : )
Hey guys, I'm back and I'm going to say the solution to my prior problem. The guy who said the answer was 47 is right. The problem is really clear though, it's asking for the number of factors 2 P has, considering P=1x2x3x4x...x50. Now we know that there are 25 pair numbers in P, but, for example, 8 has 3 factors 2, because 8=2x2x2, 18 has 1 factor 2 because 18=3x3x2. Knowing that a multiplication means the addition of the powers, then we can get all factors 2 of the numbers 1, 2, 3, ..., 50 and add them. The result is 47. ;D Great job for the one who did it!
Tyrannical said:
OK, a harder one
A * B * C = 528 A * C = 78 B * C = 84 |
Since the addition of ABC was too easy, I'm gonna post the answer to this one, using only one-step solutions
A=(528/84)==A= 44/7
B=(528/78) ==B= 88/13
C=(528/((44/7)(88/13)))==C=(273/22)/// And these are the solutions. It's actually very easy.
***
colonelstubbs...that's not actually a mathematic problem, though it's still very interesting.
***
I'd love to answer Twestern's problem, unfortunately I haven't learned limits yet. That's part of the senior school program and I'm still a junior.
Here's another problem I hope you like this one. Keep up with the problems guys! ;)
An international organization has 32 members. ¿How many members would it have in 3 years if the number of members increases every year in respect to the last in a 50%?
Answers: a)182 b)128 c)108 d)96 e)80
Picking an answer just for the hell of it won't do, you have to validate your answer! This problem is actually very easy. I'll be checking by tomorrow. Bye!
| TheLivingShadow said: Hey guys, I'm back and I'm going to say the solution to my prior problem. The guy who said the answer was 47 is right. The problem is really clear though, it's asking for the number of factors 2 P has, considering P=1x2x3x4x...x50. Now we know that there are 25 pair numbers in P, but, for example, 8 has 3 factors 2, because 8=2x2x2, 18 has 1 factor 2 because 18=3x3x2. Knowing that a multiplication means the addition of the powers, then we can get all factors 2 of the numbers 1, 2, 3, ..., 50 and add them. The result is 47. ;D Great job for the one who did it!
Since the addition of ABC was too easy, I'm gonna post the answer to this one, using only one-step solutions A=(528/84)==A= 44/7 *** *** Here's another problem I hope you like this one. Keep up with the problems guys! ;) An international organization has 32 members. ¿How many members would it have in 3 years if the number of members increases every year in respect to the last in a 50%? Answers: a)182 b)128 c)108 d)96 e)80 Picking an answer just for the hell of it won't do, you have to validate your answer! This problem is actually very easy. I'll be checking by tomorrow. Bye! |
Year 1: 32 * 1.5 = 48
Year 2: 48 * 1.5 = 72
Year 3: 72 * 1.5 = 108
Answer-- C
And the problem I asked ventures into Calculus II territory. I assume most high schools offer it since mine did (though it only ever had one student while I was there, not me) and I went to a fairly small podunk town public school.
Though if you even know what the second problem is asking, the first one doesn't involve any work (and I suppose Tyrranical is techinically right I believe, just wasn't the answer I was looking for).
Now, if people are *really* in for some nerd stuff, here are some fun math problems (not high school level):
If F(s) = s3 / (s2 + 9)2, find f(t) using Laplace transforms
Solution
s3 / (s2 + 9)2 = (As + B) / (s2 + 9)2 + (Cs + D) / (s2 + 9)
= (3): A[s / (s2 + 9)2] + B[1 / (s2 + 9)2] + C[ 1/ (s2 + 9)2] + (D / 3)(3 / (s2 + 9))
Using (1) and (2)...
L-1[s3 / (s2 + 9)2] = A((1 / (2 * 3))t sin (3t)) + B( (1/(2 * 33)) sin (3t)) - ((1 / (2 * 32)) t cost (3t))) + C cost (3t) + (D / 3) sin (3t)
multiply both sides of (3) by (s2 + 9)2 to get
s3 = As + B + (s2 + 0)(Cs + D)
S3: 1 = C,
S2: 0 = D,
S: 0 = A + 9C,
1 = s0: 0 = B + 9D.
A = -9,
B = 0,
C = 1,
D = 0.
(1): L-1[s / (ss + b2)] = (1 / 2b)t sin (bt)
(2): L-1[1 / (s2 + b2)2] = )(1 / (2b3)) sin (bt)) - ((1 / 2b2) / (t cost (bt)))
God I loved DiffEQ.
| TheLivingShadow said: Hey there guys! I was thinking, since I consider (most of) you people very smart individuals, I figured you'd be good for mathematics, like me. So then, do you have talent with numbers? I made this thread so we can discuss this very interesting area of science. If you've got any math problems that need a solution, maybe I could help, as long as it's up to high school level (no university yet hehe). To open up a conversation, have you ever participated in a math contest? I have...actually I am hehe and the winners haven't been announced yet. Suffice to say that I've cleared 2/3 rounds of tests and I'm the only guy from my province who made it to the finals; that, and the fact that I made a very good final exam. ;) |
Did you ever see any of my posts? How about RolStoppable or kirby007's? My point exactly.
Also, jesus kung fu magic
| twesterm said: Now, if people are *really* in for some nerd stuff, here are some fun math problems (not high school level): Solution s3 / (s2 + 9)2 = (As + B) / (s2 + 9)2 + (Cs + D) / (s2 + 9) = (3): A[s / (s2 + 9)2] + B[1 / (s2 + 9)2] + C[ 1/ (s2 + 9)2] + (D / 3)(3 / (s2 + 9)) Using (1) and (2)... L-1[s3 / (s2 + 9)2] = A((1 / (2 * 3))t sin (3t)) + B( (1/(2 * 33)) sin (3t)) - ((1 / (2 * 32)) t cost (3t))) + C cost (3t) + (D / 3) sin (3t) multiply both sides of (3) by (s2 + 9)2 to get s3 = As + B + (s2 + 0)(Cs + D) S3: 1 = C, A = -9, (1): L-1[s / (ss + b2)] = (1 / 2b)t sin (bt) (2): L-1[1 / (s2 + b2)2] = )(1 / (2b3)) sin (bt)) - ((1 / 2b2) / (t cost (bt))) God I loved DiffEQ.
|
You're right about my problem, good for you!
However I could never answer that now because that is beyond my current knowledge (though not perhaps capabilities). I don't even know who's Laplace...
Also, it seems that you guys have misunderstood me a little. Mathematic problems are not the same as mathematic exercises. Problems make you think, exercises make you practice knowledge. Please try to get some problems if you can.
By the way since you answered that one correctly, here's a harder one for anyone who may want to do it.
There's N which is equal to the product of the first 99 numbers. There's also M which is equal to the product of the reverse of the first 99 numbers. If a number has only one digit, its reverse is the same number. If the digit is of the form ab, its reverse is of the form ba. Calculate the value of
M/N
I shall wait for some answers!
| TheLivingShadow said: 4 You're right about my problem, good for you! However I could never answer that now because that is beyond my current knowledge (though not perhaps capabilities). I don't even know who's Laplace... Also, it seems that you guys have misunderstood me a little. Mathematic problems are not the same as mathematic exercises. Problems make you think, exercises make you practice knowledge. Please try to get some problems if you can. By the way since you answered that one correctly, here's a harder one for anyone who may want to do it. There's N which is equal to the product of the first 99 numbers. There's also M which is equal to the product of the reverse of the first 99 numbers. If a number has only one digit, its reverse is the same number. If the digit is of the form ab, its reverse is of the form ba. Calculate the value of M/N I shall wait for some answers! |
If I'm not mistaken, the answer should be 10^9. That's because all the numbers except the ones ended in zero cancel themselves, thus leaving 10,20,30...90 over 1,2...9. (Because 90 inverted leaves 09, 80 inverted leaves 08 and so on) Dividing leaves us with nine tens, thus 10^9. I hope it's correct
TheLivingShadow said:
You're right about my problem, good for you! However I could never answer that now because that is beyond my current knowledge (though not perhaps capabilities). I don't even know who's Laplace... Also, it seems that you guys have misunderstood me a little. Mathematic problems are not the same as mathematic exercises. Problems make you think, exercises make you practice knowledge. Please try to get some problems if you can. By the way since you answered that one correctly, here's a harder one for anyone who may want to do it. There's N which is equal to the product of the first 99 numbers. There's also M which is equal to the product of the reverse of the first 99 numbers. If a number has only one digit, its reverse is the same number. If the digit is of the form ab, its reverse is of the form ba. Calculate the value of M/N I shall wait for some answers! |
Sadly don't have time to think about your problem (or even read the FF thread I made ;_;), though I'm going to take a wild shot in the dark and say it's 1, but I will say this, that thing I posted above is a problem and not just an exercise.
I skipped A LOT of thinking steps in there and just stuck to the answer. You have to put a lot of thought in how you go about solving that problem and not just apply a few formulas. You have to move things around just right to get things into a specific type or else it just doesn't work.
So yeah, not really a logic problem but not just an exercise either.
Usually those types of problems take me about 1-2 pages of work (turning in homework for that class was like turning in a book).
Merry Christmas, I'm off to my parents and then the wife's parents tomorrow ^^
@TheLivingShadow
is M/N = 1
Reason: All the numbers will cancel each other out. 68 will cancel with 86 when you reverse it and 86 will cancel with 68 when you reverse it and so on.
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